Mix Examples - Number Systems Questions in English

The decimal expansions are $\sqrt{2} \approx 1.4142135 \ldots$ and $\sqrt{3} \approx 1.7320508 \ldots$
$1$. Rational Number: $A$ terminating decimal like $1.5$ lies between $1.4142135 \ldots$ and $1.7320508 \ldots$. Since $1.5 = \frac{15}{10} = \frac{3}{2}$,it is a rational number.
$2$. Irrational Number: $A$ non-terminating and non-recurring decimal like $1.575575557 \ldots$ lies between $1.4142135 \ldots$ and $1.7320508 \ldots$. Thus,$1.575575557 \ldots$ is an irrational number.

(N/A) rational number is any number that can be expressed in the form $p/q$ where $p$ and $q$ are integers and $q \neq 0$. Since $2.357 < 3 < 3.121$,the integer $3$ is a rational number between $2.357$ and $3.121$.
An irrational number is a number whose decimal expansion is non-terminating and non-recurring. $A$ number like $3.101101110\dots$ satisfies $2.357 < 3.101101110\dots < 3.121$,therefore,it is an irrational number between the given values.

(N/A) rational number is a number that can be expressed in the form $p/q$,where $p$ and $q$ are integers and $q \neq 0$. $A$ terminating decimal is a rational number.
$1$. Rational number: We can choose $0.0005$. Since $0.0005 = 5/10000 = 1/2000$,it is a rational number lying between $0.0001$ and $0.001$.
$2$. Irrational number: An irrational number is a non-terminating and non-recurring decimal. We can construct one by following a pattern: $0.0001010010001...$ This number is clearly greater than $0.0001$ and less than $0.001$,and it does not terminate or repeat,making it an irrational number.

(N/A) To find a rational number between $0.484848$ and $3.623623$,we can choose any terminating decimal or integer within this range. For example,$1$ is a rational number because it can be expressed as $\frac{1}{1}$.
To find an irrational number between $0.484848$ and $3.623623$,we need a non-terminating and non-recurring decimal. For example,$1.909009000\dots$ is an irrational number because its decimal expansion neither terminates nor repeats.

(N/A) $6.3753$ (a terminating decimal) is a rational number between $6.375289$ and $6.375738$.
$6.375414114111...$ (a non-terminating and non-recurring decimal) is an irrational number lying between $6.375289$ and $6.375738$.

(N/A) To represent the given numbers on the number line:
$1$. $7$: This is a positive integer. Locate $7$ on the number line to the right of $0$.
$2$. $7.2$: This is $7 + 0.2$. Divide the segment between $7$ and $8$ into $10$ equal parts. The second mark after $7$ represents $7.2$.
$3$. $\frac{-3}{2} = -1.5$: This is a negative rational number. It lies exactly halfway between $-1$ and $-2$ on the left side of $0$.
$4$. $\frac{-12}{5} = -2.4$: This is a negative rational number. Divide the segment between $-2$ and $-3$ into $5$ equal parts. The fourth mark to the left of $-2$ represents $-2.4$.

(N/A) Presentation of $\sqrt{5}$ on the number line:
We write $5$ as the sum of the squares of two natural numbers:
$5 = 1 + 4 = 1^{2} + 2^{2}$
On the number line,take $OA = 2$ units.
Draw $BA = 1$ unit,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{2^{2} + 1^{2}} = \sqrt{5}$.
Using a compass with center $O$ and radius $OB$,draw an arc which intersects the number line at point $C$. Then,$C$ corresponds to $\sqrt{5}$.
Presentation of $\sqrt{10}$ on the number line:
We write $10$ as the sum of the squares of two natural numbers:
$10 = 1 + 9 = 1^{2} + 3^{2}$
On the number line,take $OA = 3$ units.
Draw $BA = 1$ unit,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{3^{2} + 1^{2}} = \sqrt{10}$.
Using a compass with center $O$ and radius $OB$,draw an arc which intersects the number line at point $C$. Then,$C$ corresponds to $\sqrt{10}$.
Presentation of $\sqrt{17}$ on the number line:
We write $17$ as the sum of the squares of two natural numbers:
$17 = 1 + 16 = 1^{2} + 4^{2}$
On the number line,take $OA = 4$ units.
Draw $BA = 1$ unit,perpendicular to $OA$. Join $OB$.
By Pythagoras theorem,$OB = \sqrt{4^{2} + 1^{2}} = \sqrt{17}$.
Using a compass with center $O$ and radius $OB$,draw an arc which intersects the number line at point $C$. Then,$C$ corresponds to $\sqrt{17}$.

(N/A) $1$. Draw a line segment $AB = 4.5$ units on a number line.
$2$. From point $B$,mark a distance of $1$ unit and label the new point as $C$. Now,$AC = 4.5 + 1 = 5.5$ units.
$3$. Find the midpoint of $AC$ and label it as $O$.
$4$. Draw a semicircle with center $O$ and radius $OC$ (or $OA$).
$5$. Draw a line perpendicular to $AC$ at point $B$,which intersects the semicircle at point $D$. The length of $BD$ is $\sqrt{4.5}$.
$6$. With $B$ as the center and $BD$ as the radius,draw an arc that intersects the number line at point $E$. The point $E$ represents $\sqrt{4.5}$ on the number line.

(N/A) To represent $\sqrt{5.6}$ on the number line,follow these steps:
$1$. Draw a line segment $AB = 5.6 \text{ units}$ on the number line.
$2$. From point $B$,mark a point $C$ such that $BC = 1 \text{ unit}$. Now,$AC = 5.6 + 1 = 6.6 \text{ units}$.
$3$. Find the midpoint $O$ of $AC$. The length $AO = OC = 6.6 / 2 = 3.3 \text{ units}$.
$4$. Draw a semicircle with center $O$ and radius $OA = 3.3 \text{ units}$.
$5$. Draw a perpendicular line at point $B$ to the line $AC$,which intersects the semicircle at point $D$.
$6$. The length $BD$ is equal to $\sqrt{5.6}$.
$7$. With $B$ as the center and $BD$ as the radius,draw an arc on the number line to intersect it at point $E$. The distance $BE$ represents $\sqrt{5.6}$ on the number line.

(N/A) $1$. Draw a line segment $AB = 8.1$ units on a number line.
$2$. From point $B$,mark a distance of $1$ unit and label the new point as $C$. Now,$AC = 8.1 + 1 = 9.1$ units.
$3$. Find the midpoint of $AC$ and label it as $O$.
$4$. Draw a semicircle with center $O$ and radius $OC$ (or $OA$).
$5$. Draw a line perpendicular to $AC$ at point $B$,which intersects the semicircle at point $D$. The length of $BD$ is $\sqrt{8.1}$.
$6$. With $B$ as the center and $BD$ as the radius,draw an arc that intersects the number line at point $E$. The distance $BE$ represents $\sqrt{8.1}$ on the number line.

(N/A) $1$. Draw a line segment $AB = 2.3$ units on a line.
$2$. From $B$,mark a distance of $1$ unit and label the new point as $C$. Now,$AC = 2.3 + 1 = 3.3$ units.
$3$. Find the midpoint of $AC$ and mark it as $O$.
$4$. Draw a semicircle with center $O$ and radius $OC$.
$5$. Draw a line perpendicular to $AC$ passing through $B$ that intersects the semicircle at point $D$. The length $BD$ is equal to $\sqrt{2.3}$.
$6$. With $B$ as the center and $BD$ as the radius,draw an arc that intersects the number line at point $E$. The point $E$ represents $\sqrt{2.3}$ on the number line.

(B) To express the decimal $0.2$ in the form $\frac{p}{q}$,we follow these steps:
$1$. Write the decimal as a fraction with a denominator of $10$ because there is one digit after the decimal point: $0.2 = \frac{2}{10}$.
$2$. Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor,which is $2$: $\frac{2 \div 2}{10 \div 2} = \frac{1}{5}$.
$3$. Thus,$0.2 = \frac{1}{5}$,where $p = 1$ and $q = 5$ are integers and $q \neq 0$.

(C) Let $x = 0.888 \ldots = 0.\overline{8} \quad \dots(1)$
Multiplying both sides by $10$,we get:
$10x = 8.888 \ldots = 8.\overline{8} \quad \dots(2)$
Subtracting equation $(1)$ from equation $(2)$:
$10x - x = 8.\overline{8} - 0.\overline{8}$
$9x = 8$
Therefore,$x = \frac{8}{9}$.

(D) Let $x = 5. \overline{2} = 5.2222 \ldots \quad \dots(1)$
Multiplying both sides by $10,$ we get
$10x = 52.2222 \ldots = 52. \overline{2} \quad \dots(2)$
Subtracting equation $(1)$ from equation $(2)$:
$10x - x = 52.2222 \ldots - 5.2222 \ldots$
$9x = 47$
$x = \frac{47}{9}$
Hence,$5. \overline{2} = \frac{47}{9}$.

(A) Let $x = 0.\overline{001} = 0.001001...$ $(1)$
Since there are three repeating digits,multiply both sides by $1000$:
$1000x = 1.001001...$ $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$1000x - x = 1.001001... - 0.001001...$
$999x = 1$
$x = \frac{1}{999}$
Thus,the number $0.\overline{001}$ in the form $\frac{p}{q}$ is $\frac{1}{999}$.

(B) Let $x = 0.2555 \ldots = 0.2\overline{5} \quad ....(1)$
Multiply both sides by $10$:
$10x = 2.555 \ldots = 2.\overline{5} \quad ....(2)$
Multiply both sides by $100$:
$100x = 25.555 \ldots = 25.\overline{5} \quad ....(3)$
Subtracting equation $(2)$ from equation $(3)$:
$100x - 10x = 25.\overline{5} - 2.\overline{5}$
$90x = 23$
Therefore,$x = \frac{23}{90}$.

(C) Let $x = 0.1\overline{34} = 0.1343434... \quad ....(1)$
Multiply both sides by $10$ to shift the non-repeating part:
$10x = 1.343434... \quad ....(2)$
Multiply equation $(2)$ by $100$ to shift the repeating part:
$1000x = 134.343434... \quad ....(3)$
Subtract equation $(2)$ from equation $(3)$:
$1000x - 10x = 134.343434... - 1.343434...$
$990x = 133$
Therefore,$x = \frac{133}{990}$.

(D) Let $x = 0.00323232 \ldots = 0.00\overline{32} \quad \dots(1)$
Multiply equation $(1)$ by $100$ to shift the decimal point:
$100x = 0.323232 \ldots = 0.\overline{32} \quad \dots(2)$
Multiply equation $(2)$ by $100$ again to shift the decimal point past the repeating part:
$10000x = 32.323232 \ldots = 32.\overline{32} \quad \dots(3)$
Subtracting equation $(2)$ from equation $(3)$:
$10000x - 100x = 32.\overline{32} - 0.\overline{32}$
$9900x = 32$
$x = \frac{32}{9900}$
Simplifying the fraction by dividing both numerator and denominator by $4$:
$x = \frac{32 \div 4}{9900 \div 4} = \frac{8}{2475}$

(A) Let $x = 0.404040 \ldots = 0.\overline{40} \quad \dots(1)$
Since there are two repeating digits,multiply both sides by $100$:
$100x = 40.404040 \ldots = 40.\overline{40} \quad \dots(2)$
Subtracting equation $(1)$ from equation $(2)$:
$100x - x = 40.\overline{40} - 0.\overline{40}$
$99x = 40$
$x = \frac{40}{99}$
Thus,the number $0.404040 \ldots$ expressed in the form $\frac{p}{q}$ is $\frac{40}{99}$.

(N/A) Let $x = 0.142857142857 \ldots$ $(1)$
Since the repeating block has $6$ digits,multiply both sides by $1,000,000$:
$1,000,000 x = 142857.142857142857 \ldots$ $(2)$
Subtracting equation $(1)$ from equation $(2)$:
$1,000,000 x - x = 142857.142857 \ldots - 0.142857 \ldots$
$999,999 x = 142857$
Solving for $x$:
$x = \frac{142857}{999999}$
Dividing both numerator and denominator by $142857$:
$x = \frac{1}{7}$
Thus,$0.142857142857 \ldots = \frac{1}{7}$.

(C) To simplify the expression $\sqrt{45}-3 \sqrt{20}+4 \sqrt{5}$,we first factor the numbers inside the square roots:
$\sqrt{45} = \sqrt{9 \times 5} = 3 \sqrt{5}$
$\sqrt{20} = \sqrt{4 \times 5} = 2 \sqrt{5}$
Now,substitute these values back into the expression:
$3 \sqrt{5} - 3(2 \sqrt{5}) + 4 \sqrt{5}$
$= 3 \sqrt{5} - 6 \sqrt{5} + 4 \sqrt{5}$
$= (3 - 6 + 4) \sqrt{5}$
$= 1 \sqrt{5} = \sqrt{5}$

(D) First,simplify the square roots:
$\sqrt{24} = \sqrt{4 \times 6} = 2 \sqrt{6}$
$\sqrt{54} = \sqrt{9 \times 6} = 3 \sqrt{6}$
Substitute these into the expression:
$\frac{2 \sqrt{6}}{8} + \frac{3 \sqrt{6}}{9} = \frac{\sqrt{6}}{4} + \frac{\sqrt{6}}{3}$
Find a common denominator,which is $12$:
$\frac{3 \sqrt{6}}{12} + \frac{4 \sqrt{6}}{12} = \frac{3 \sqrt{6} + 4 \sqrt{6}}{12} = \frac{7 \sqrt{6}}{12}$

(A) To simplify the expression $4 \sqrt{12} \times 7 \sqrt{6}$,we first factorize the numbers inside the square roots:
$4 \sqrt{12} = 4 \sqrt{4 \times 3} = 4 \times 2 \sqrt{3} = 8 \sqrt{3}$
$7 \sqrt{6} = 7 \sqrt{2 \times 3}$
Now,multiply the two expressions:
$8 \sqrt{3} \times 7 \sqrt{2 \times 3} = (8 \times 7) \times (\sqrt{3} \times \sqrt{3}) \times \sqrt{2}$
$= 56 \times 3 \times \sqrt{2}$
$= 168 \sqrt{2}$

(B) To simplify the expression $4 \sqrt{28} \div 3 \sqrt{7}$,we first express the division as a fraction:
$\frac{4 \sqrt{28}}{3 \sqrt{7}}$
Next,we simplify the square root of $28$ by factoring it as $28 = 4 \times 7$:
$\sqrt{28} = \sqrt{4 \times 7} = 2 \sqrt{7}$
Now,substitute this back into the expression:
$\frac{4 \times (2 \sqrt{7})}{3 \sqrt{7}}$
Cancel the common term $\sqrt{7}$ from the numerator and the denominator:
$\frac{4 \times 2}{3} = \frac{8}{3}$

(B) Given expression: $3 \sqrt{3}+2 \sqrt{27}+\frac{7}{\sqrt{3}}$
First,simplify $\sqrt{27}$:
$\sqrt{27} = \sqrt{9 \times 3} = 3 \sqrt{3}$
Substitute this into the expression:
$=3 \sqrt{3}+2(3 \sqrt{3})+\frac{7}{\sqrt{3}}$
$=3 \sqrt{3}+6 \sqrt{3}+\frac{7}{\sqrt{3}}$
Rationalize the denominator of the third term:
$\frac{7}{\sqrt{3}} = \frac{7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{7 \sqrt{3}}{3}$
Now add the terms:
$=3 \sqrt{3}+6 \sqrt{3}+\frac{7 \sqrt{3}}{3}$
$=9 \sqrt{3}+\frac{7 \sqrt{3}}{3}$
$= \sqrt{3} \left(9+\frac{7}{3}\right)$
$= \sqrt{3} \left(\frac{27+7}{3}\right)$
$= \frac{34}{3} \sqrt{3}$

(D) To simplify the expression $(\sqrt{3}-\sqrt{2})^{2}$,we use the algebraic identity $(a-b)^{2} = a^{2} + b^{2} - 2ab$.
Here,$a = \sqrt{3}$ and $b = \sqrt{2}$.
Substituting these values into the identity:
$(\sqrt{3}-\sqrt{2})^{2} = (\sqrt{3})^{2} + (\sqrt{2})^{2} - 2(\sqrt{3})(\sqrt{2})$
Since $(\sqrt{x})^{2} = x$ and $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$:
$= 3 + 2 - 2\sqrt{3 \times 2}$
$= 5 - 2\sqrt{6}$

(A) To simplify the expression $\sqrt[4]{81}-8 \sqrt[3]{216}+15 \sqrt[5]{32}+\sqrt{225}$,we evaluate each radical term individually:
$1$. $\sqrt[4]{81} = \sqrt[4]{3^{4}} = 3$
$2$. $\sqrt[3]{216} = \sqrt[3]{6^{3}} = 6$,so $8 \sqrt[3]{216} = 8 \times 6 = 48$
$3$. $\sqrt[5]{32} = \sqrt[5]{2^{5}} = 2$,so $15 \sqrt[5]{32} = 15 \times 2 = 30$
$4$. $\sqrt{225} = \sqrt{15^{2}} = 15$
Substituting these values back into the original expression:
$3 - 48 + 30 + 15$
$= (3 + 30 + 15) - 48$
$= 48 - 48 = 0$

(B) Given expression: $\frac{3}{\sqrt{8}} + \frac{1}{\sqrt{2}}$
First,simplify $\sqrt{8}$ as $\sqrt{4 \times 2} = 2\sqrt{2}$.
Substitute this into the expression: $\frac{3}{2\sqrt{2}} + \frac{1}{\sqrt{2}}$.
To add these fractions,find a common denominator,which is $2\sqrt{2}$.
$\frac{3}{2\sqrt{2}} + \frac{1 \times 2}{\sqrt{2} \times 2} = \frac{3}{2\sqrt{2}} + \frac{2}{2\sqrt{2}}$.
$= \frac{3 + 2}{2\sqrt{2}} = \frac{5}{2\sqrt{2}}$.
Rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$:
$= \frac{5 \times \sqrt{2}}{2\sqrt{2} \times \sqrt{2}} = \frac{5\sqrt{2}}{2 \times 2} = \frac{5\sqrt{2}}{4}$.

(C) To simplify the expression $\frac{2 \sqrt{3}}{3} - \frac{\sqrt{3}}{6}$,we first find a common denominator for the fractions,which is $6$.
$\frac{2 \sqrt{3}}{3} - \frac{\sqrt{3}}{6} = \frac{2 \sqrt{3} \times 2}{3 \times 2} - \frac{\sqrt{3}}{6}$
$= \frac{4 \sqrt{3}}{6} - \frac{\sqrt{3}}{6}$
$= \frac{4 \sqrt{3} - \sqrt{3}}{6}$
$= \frac{3 \sqrt{3}}{6}$
$= \frac{\sqrt{3}}{2}$

(D) To rationalise the denominator of $\frac{2}{3 \sqrt{3}}$,we multiply both the numerator and the denominator by $\sqrt{3}$:
$\frac{2}{3 \sqrt{3}} = \frac{2}{3 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{2 \sqrt{3}}{3 \times 3}$
$= \frac{2 \sqrt{3}}{9}$

(A) To rationalise the denominator of $\frac{\sqrt{40}}{\sqrt{3}}$,we multiply both the numerator and the denominator by $\sqrt{3}$.
$\frac{\sqrt{40}}{\sqrt{3}} = \frac{\sqrt{40} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$
Since $\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$,we have:
$= \frac{2\sqrt{10} \times \sqrt{3}}{3}$
$= \frac{2\sqrt{30}}{3}$

(B) To rationalise the denominator of $\frac{3+\sqrt{2}}{4 \sqrt{2}}$,we multiply both the numerator and the denominator by $\sqrt{2}$:
$\frac{3+\sqrt{2}}{4 \sqrt{2}} = \frac{3+\sqrt{2}}{4 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
$= \frac{\sqrt{2}(3+\sqrt{2})}{4 \times (\sqrt{2} \times \sqrt{2})}$
$= \frac{3 \sqrt{2} + 2}{4 \times 2}$
$= \frac{3 \sqrt{2} + 2}{8}$

(C) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{41}+5)$.
$\frac{16}{\sqrt{41}-5} = \frac{16}{\sqrt{41}-5} \times \frac{\sqrt{41}+5}{\sqrt{41}+5}$
Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator:
$= \frac{16(\sqrt{41}+5)}{(\sqrt{41})^2 - (5)^2}$
$= \frac{16(\sqrt{41}+5)}{41 - 25}$
$= \frac{16(\sqrt{41}+5)}{16}$
$= \sqrt{41}+5$

(D) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(2+\sqrt{3})$.
$\frac{2+\sqrt{3}}{2-\sqrt{3}} = \frac{2+\sqrt{3}}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}}$
Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator and $(a+b)^2 = a^2 + b^2 + 2ab$ in the numerator:
$= \frac{(2+\sqrt{3})^2}{2^2 - (\sqrt{3})^2}$
$= \frac{2^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3})}{4 - 3}$
$= \frac{4 + 3 + 4\sqrt{3}}{1}$
$= 7 + 4\sqrt{3}$

(A) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{2}-\sqrt{3})$:
$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} = \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{(\sqrt{2})^2 - (\sqrt{3})^2}$
$= \frac{\sqrt{12} - \sqrt{18}}{2 - 3}$
$= \frac{\sqrt{12} - \sqrt{18}}{-1}$
$= \sqrt{18} - \sqrt{12}$
$= \sqrt{9 \times 2} - \sqrt{4 \times 3}$
$= 3\sqrt{2} - 2\sqrt{3}$

(B) To rationalise the denominator,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{3}+\sqrt{2})$.
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
Using the identity $(a-b)(a+b) = a^2 - b^2$ in the denominator and $(a+b)^2 = a^2 + b^2 + 2ab$ in the numerator:
$= \frac{(\sqrt{3}+\sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2}$
$= \frac{3 + 2 + 2\sqrt{3} \times \sqrt{2}}{3 - 2}$
$= \frac{5 + 2\sqrt{6}}{1}$
$= 5 + 2\sqrt{6}$

(C) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{5}+\sqrt{3})$.
$\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$
$= \frac{3(\sqrt{5})^2 + 3\sqrt{15} + \sqrt{15} + (\sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2}$
$= \frac{3(5) + 4\sqrt{15} + 3}{5 - 3}$
$= \frac{15 + 4\sqrt{15} + 3}{2}$
$= \frac{18 + 4\sqrt{15}}{2}$
$= \frac{2(9 + 2\sqrt{15})}{2}$
$= 9 + 2\sqrt{15}$

(D) Given expression: $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$
Simplify the denominator: $\sqrt{48} = \sqrt{16 \times 3} = 4 \sqrt{3}$ and $\sqrt{18} = \sqrt{9 \times 2} = 3 \sqrt{2}$.
So,the expression becomes: $\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}}$.
To rationalise the denominator,multiply the numerator and denominator by the conjugate $(4 \sqrt{3}-3 \sqrt{2})$:
$= \frac{(4 \sqrt{3}+5 \sqrt{2})(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3}+3 \sqrt{2})(4 \sqrt{3}-3 \sqrt{2})}$
$= \frac{(4 \sqrt{3})^2 - (4 \sqrt{3})(3 \sqrt{2}) + (5 \sqrt{2})(4 \sqrt{3}) - (5 \sqrt{2})(3 \sqrt{2})}{(4 \sqrt{3})^2 - (3 \sqrt{2})^2}$
$= \frac{48 - 12 \sqrt{6} + 20 \sqrt{6} - 30}{48 - 18}$
$= \frac{18 + 8 \sqrt{6}}{30}$
Divide numerator and denominator by $2$:
$= \frac{9 + 4 \sqrt{6}}{15}$

(A) To solve the expression,we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator,which is $(7-4 \sqrt{3})$.
$L.H.S. = \frac{5+2 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}$
$= \frac{(5+2 \sqrt{3})(7-4 \sqrt{3})}{(7)^2 - (4 \sqrt{3})^2}$
$= \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 24}{49 - 48}$
$= \frac{11 - 6 \sqrt{3}}{1} = 11 - 6 \sqrt{3}$
Given that $11 - 6 \sqrt{3} = a - 6 \sqrt{3}$,by comparing both sides,we get $a = 11$.

(NONE) To solve the expression,we rationalize the denominator:
$L.H.S. = \frac{3-\sqrt{5}}{3+2 \sqrt{5}} \times \frac{3-2 \sqrt{5}}{3-2 \sqrt{5}}$
$= \frac{9 - 6\sqrt{5} - 3\sqrt{5} + 2(5)}{3^2 - (2\sqrt{5})^2}$
$= \frac{9 - 9\sqrt{5} + 10}{9 - 20} = \frac{19 - 9\sqrt{5}}{-11}$
$= -\frac{19}{11} + \frac{9}{11}\sqrt{5}$
Comparing this with $a\sqrt{5} - \frac{19}{11}$,we get:
$a\sqrt{5} - \frac{19}{11} = \frac{9}{11}\sqrt{5} - \frac{19}{11}$
Therefore,$a = \frac{9}{11}$.

(C) Given expression: $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}} = 2 - b\sqrt{6}$
Rationalize the denominator by multiplying the numerator and denominator by $(3\sqrt{2} + 2\sqrt{3})$:
$= \frac{(\sqrt{2}+\sqrt{3})(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})}$
$= \frac{3(\sqrt{2})^2 + 2\sqrt{6} + 3\sqrt{6} + 2(\sqrt{3})^2}{(3\sqrt{2})^2 - (2\sqrt{3})^2}$
$= \frac{3(2) + 5\sqrt{6} + 2(3)}{18 - 12}$
$= \frac{6 + 5\sqrt{6} + 6}{6} = \frac{12 + 5\sqrt{6}}{6} = 2 + \frac{5}{6}\sqrt{6}$
Comparing $2 + \frac{5}{6}\sqrt{6}$ with $2 - b\sqrt{6}$:
$-b = \frac{5}{6} \Rightarrow b = -\frac{5}{6}$

(D) Given equation: $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} \sqrt{5} b$
Rationalizing the denominators:
$\frac{(7+\sqrt{5})^2}{(7-\sqrt{5})(7+\sqrt{5})} - \frac{(7-\sqrt{5})^2}{(7+\sqrt{5})(7-\sqrt{5})} = a+\frac{7}{11} \sqrt{5} b$
Using $(a+b)^2 = a^2+b^2+2ab$ and $(a-b)^2 = a^2+b^2-2ab$:
$\frac{49+5+14\sqrt{5}}{49-5} - \frac{49+5-14\sqrt{5}}{49-5} = a+\frac{7}{11} \sqrt{5} b$
$\frac{54+14\sqrt{5}}{44} - \frac{54-14\sqrt{5}}{44} = a+\frac{7}{11} \sqrt{5} b$
$\frac{54+14\sqrt{5}-54+14\sqrt{5}}{44} = a+\frac{7}{11} \sqrt{5} b$
$\frac{28\sqrt{5}}{44} = a+\frac{7}{11} \sqrt{5} b$
Simplifying the fraction $\frac{28}{44}$ by dividing by $4$:
$\frac{7\sqrt{5}}{11} = a+\frac{7}{11} \sqrt{5} b$
Comparing both sides,we get $a=0$ and $b=1$.

(A) Given that $a = 2 + \sqrt{3}$.
First,we find the value of $\frac{1}{a}$ by rationalizing the denominator:
$\frac{1}{a} = \frac{1}{2 + \sqrt{3}}$
Multiply the numerator and denominator by the conjugate $(2 - \sqrt{3})$:
$\frac{1}{a} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}}$
Using the identity $(x + y)(x - y) = x^2 - y^2$ in the denominator:
$\frac{1}{a} = \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2}$
$\frac{1}{a} = \frac{2 - \sqrt{3}}{4 - 3} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}$
Now,calculate $a - \frac{1}{a}$:
$a - \frac{1}{a} = (2 + \sqrt{3}) - (2 - \sqrt{3})$
$a - \frac{1}{a} = 2 + \sqrt{3} - 2 + \sqrt{3}$
$a - \frac{1}{a} = 2\sqrt{3}$

(B) To rationalise the denominator,multiply the numerator and the denominator by $\sqrt{3}$:
$\frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{4\sqrt{3}}{3}$
Substitute the value $\sqrt{3} = 1.732$:
$= \frac{4 \times 1.732}{3}$
$= \frac{6.928}{3}$
$= 2.309$

(C) To rationalise the denominator of $\frac{6}{\sqrt{6}}$,we multiply the numerator and the denominator by $\sqrt{6}$:
$\frac{6}{\sqrt{6}} = \frac{6 \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}} = \frac{6\sqrt{6}}{6} = \sqrt{6}$
We know that $\sqrt{6} = \sqrt{2 \times 3} = \sqrt{2} \times \sqrt{3}$.
Given $\sqrt{2} = 1.414$ and $\sqrt{3} = 1.732$,we have:
$\sqrt{6} = 1.414 \times 1.732 = 2.449048$
Rounding to three decimal places,we get $2.449$.

(D) Given expression: $\frac{\sqrt{10}-\sqrt{5}}{2}$
Step $1$: Simplify the numerator by factoring out $\sqrt{5}$:
$\frac{\sqrt{5} \times \sqrt{2} - \sqrt{5}}{2} = \frac{\sqrt{5}(\sqrt{2}-1)}{2}$
Step $2$: Substitute the given values $\sqrt{5} = 2.236$ and $\sqrt{2} = 1.414$:
$= \frac{2.236(1.414 - 1)}{2}$
Step $3$: Perform the subtraction inside the parenthesis:
$= \frac{2.236(0.414)}{2}$
Step $4$: Divide $2.236$ by $2$:
$= 1.118 \times 0.414$
Step $5$: Multiply the values:
$= 0.462852$
Rounding to three decimal places,we get $0.463$.

(A) To rationalise the denominator of $\frac{\sqrt{2}}{2+\sqrt{2}}$,we multiply the numerator and the denominator by the conjugate of the denominator,which is $(2-\sqrt{2})$.
$\frac{\sqrt{2}}{2+\sqrt{2}} = \frac{\sqrt{2}}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{\sqrt{2}(2-\sqrt{2})}{(2)^2 - (\sqrt{2})^2}$
$= \frac{2\sqrt{2} - 2}{4 - 2}$
$= \frac{2\sqrt{2} - 2}{2}$
$= \frac{2(\sqrt{2} - 1)}{2}$
$= \sqrt{2} - 1$
Given $\sqrt{2} = 1.414$,we substitute the value:
$= 1.414 - 1 = 0.414$

(B) To rationalise the denominator,multiply the numerator and the denominator by the conjugate of the denominator,which is $(\sqrt{3}-\sqrt{2})$.
$\frac{1}{\sqrt{3}+\sqrt{2}} = \frac{1}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
Using the identity $(a+b)(a-b) = a^2 - b^2$ in the denominator:
$= \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}$
$= \frac{\sqrt{3}-\sqrt{2}}{3-2}$
$= \frac{\sqrt{3}-\sqrt{2}}{1} = \sqrt{3}-\sqrt{2}$
Now,substitute the given values $\sqrt{3} = 1.732$ and $\sqrt{2} = 1.414$:
$= 1.732 - 1.414$
$= 0.318$

(B) Step $1$: Calculate the cubes of the numbers inside the parentheses.
$1^{3} = 1 \times 1 \times 1 = 1$
$2^{3} = 2 \times 2 \times 2 = 8$
$3^{3} = 3 \times 3 \times 3 = 27$
Step $2$: Add the results together.
$1 + 8 + 27 = 36$
Step $3$: Apply the exponent $\frac{1}{2}$,which represents the square root.
$(36)^{\frac{1}{2}} = \sqrt{36} = 6$
Therefore,the simplified value is $6$.

(25/9) To simplify the expression $(\frac{3}{5})^4 \times (\frac{3}{5})^{-12} \times (\frac{3}{5})^{6}$,we use the law of exponents: $a^m \times a^n \times a^p = a^{m+n+p}$.
Here,the base $a = \frac{3}{5}$,$m = 4$,$n = -12$,and $p = 6$.
Applying the rule: $(\frac{3}{5})^{4 + (-12) + 6}$.
Calculating the exponent: $4 - 12 + 6 = -2$.
So,the expression becomes $(\frac{3}{5})^{-2}$.
Using the rule $a^{-n} = (\frac{1}{a})^n$,we get $(\frac{5}{3})^2$.
Calculating the final value: $(\frac{5}{3})^2 = \frac{25}{9}$.