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(D) The given inequality is $\frac{x^4 - 4x^3 + 3x^2}{(x^2 - 4)(x^2 - 7x + 10)} \ge 0$. Factorizing the numerator: $x^2(x^2 - 4x + 3) = x^2(x - 1)(x -

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$( - \infty, -2 ) \cup [1, 2) \cup (2, 3] \cup (5, \infty) \cup \{0\}$

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(D) The given inequality is $\frac{x^4 - 4x^3 + 3x^2}{(x^2 - 4)(x^2 - 7x + 10)} \ge 0$. Factorizing the numerator: $x^2(x^2 - 4x + 3) = x^2(x - 1)(x - 3)$. Factorizing the denominator: $(x - 2)(x + 2)(x - 2)(x - 5) = (x + 2)(x - 2)^2(x - 5)$. The inequality becomes $\frac{x^2(x - 1)(x - 3)}{(x + 2)(x - 2)^2(x - 5)} \ge 0$. The critical points are $x = -2, 0, 1, 2, 3, 5$. Note that $x 
eq -2, 2, 5$ as they make the denominator zero. At $x = 0$,the expression is $0$,which satisfies the inequality. Using the wavy curve method (sign scheme): For $x > 5$,the expression is positive. For $3 For $2 For $1 For $0 For $-2 For $x Combining these intervals and including $x=0$,we get $( - \infty, -2 ) \cup \{0\} \cup [1, 2) \cup (2, 3] \cup (5, \infty)$.

Find the set of all values of $x$ satisfying $\frac{x^4 - 4x^3 + 3x^2}{(x^2 - 4)(x^2 - 7x + 10)} \ge 0$.

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