Solution of quadratic inequations and Newton Formula Questions in English

(D) Given inequalities are:
$1$) $x^2 - 3x - 10 < 0$
$(x - 5)(x + 2) < 0$
This implies $x \in (-2, 5)$.
$2$) $10x - x^2 - 16 > 0$
Multiply by $-1$ (inequality sign reverses):
$x^2 - 10x + 16 < 0$
$(x - 2)(x - 8) < 0$
This implies $x \in (2, 8)$.
To find the values of $x$ that satisfy both inequalities simultaneously,we take the intersection of the two sets:
$x \in (-2, 5) \cap (2, 8) = (2, 5)$.
Therefore,the correct option is $D$.

(B) Given $0 \leq p \leq 1$ and $a, b > 0$.
Consider the function $f(x) = x^p$.
Since $0 \leq p \leq 1$,the function $f(x) = x^p$ is a concave function for $x > 0$.
By the property of concave functions,for any $a, b > 0$ and $0 < p < 1$,we have $(a+b)^p \leq a^p + b^p$.
This implies $I(p) \leq J(p)$.
For $p=0$,$I(0) = (a+b)^0 = 1$ and $J(0) = a^0 + b^0 = 1 + 1 = 2$,so $1 \leq 2$.
For $p=1$,$I(1) = a+b$ and $J(1) = a+b$,so $a+b \leq a+b$.
Thus,$I(p) \leq J(p)$ holds for all $0 \leq p \leq 1$.

(D) We have,$\frac{5x^{2}-26x+5}{3x^{2}-10x+3} < 0$
Factorizing the numerator and denominator:
$\frac{5x^{2}-25x-x+5}{3x^{2}-9x-x+3} < 0$
$\frac{5x(x-5)-1(x-5)}{3x(x-3)-1(x-3)} < 0$
$\frac{(x-5)(5x-1)}{(x-3)(3x-1)} < 0$
The critical points are $x = \frac{1}{5}, \frac{1}{3}, 3, 5$.
Using the wavy curve method (sign scheme) for the expression $f(x) = \frac{(x-5)(5x-1)}{(x-3)(3x-1)}$,the expression is negative in the intervals $(\frac{1}{5}, \frac{1}{3})$ and $(3, 5)$.
Therefore,$x \in (\frac{1}{5}, \frac{1}{3}) \cup (3, 5)$.