Giải các bất phương trình `a) x^3 − 9x^2 + 26x − 24 < 0` `b)2x^3 − 6x^2 + 12x − 8 ≥ 0` `c) x^4 + x^3 − 2x^2 − 6x − 4 ≥ 0`

Đáp án:

$a)x\in (-\infty;2) \cup (3;4)\\b) x\ge 1\\ c)x\in (-\infty;1] \cup [2;+\infty).$

Giải thích các bước giải:

$a)x^3-9x^2+26x-24 <0\\ \Leftrightarrow x^3-2x^2-7x^2+14x+12x-24 <0\\ \Leftrightarrow x^2(x-2)-7x(x-2)+12(x-2) <0\\ \Leftrightarrow (x-2)(x^2-7x+12) <0\\ \Leftrightarrow (x-2)(x^2-3x-4x+12) <0\\ \Leftrightarrow (x-2)[x(x-3)-4(x-3)] <0\\ \Leftrightarrow (x-2)(x-3)(x-4) <0(1)$

BXD:

\begin{array}{|c|cc|} \hline x&-\infty&&2&&3&&4&&+\infty\\\hline x-2&&-&0&+&|&+&|&+ &\\\hline x-3 &&-&|&-&0&+&|&+& \\\hline x-4&&-&|&-&|&-&0&+&\\\hline (x-2)(x-3)(x-4) &&-&0&+&0&-&0&+&\\\hline \end{array}

Dựa vào BXD $\Rightarrow$ Nghiệm của $(1): \Leftrightarrow x\in (-\infty;2) \cup (3;4)$

$b)2x^3-6x^2+12x-8 \ge 0\\ \Leftrightarrow x^3-3x^2+6x-4 \ge 0\\ \Leftrightarrow x^3-x^2-2x^2+2x+4x-4 \ge 0\\ \Leftrightarrow x^2(x-1)-2x(x-1)+4(x-1) \ge 0\\ \Leftrightarrow (x-1)(x^2-2x+4) \ge 0\\ \Leftrightarrow (x-1)(x^2-2x+1+3) \ge 0\\ \Leftrightarrow (x-1)[(x-1)^2+3] \ge 0\\ \Leftrightarrow x-1\ge 0 (\text{Do } (x-1)^2+3>0 \ \forall \ x)\\ \Leftrightarrow x\ge 1\\ c)x^4+x^3-2x^2-6x-4 \ge 0\\ \Leftrightarrow x^4+x^3-2x^2-2x-4x-4 \ge 0\\ \Leftrightarrow x^3(x+1)-2x(x+1)-4(x+1) \ge 0\\ \Leftrightarrow (x+1)(x^3-2x-4) \ge 0\\ \Leftrightarrow (x+1)(x^3-2x^2+2x^2-4x+2x-4) \ge 0\\ \Leftrightarrow (x+1)[x^2(x-2)+2x(x-2)+2(x-2)] \ge 0\\ \Leftrightarrow (x+1)(x-2)[x^2+2x+2] \ge 0\\ \Leftrightarrow (x+1)(x-2)[x^2+2x+1+1] \ge 0\\ \Leftrightarrow (x+1)(x-2)[(x+1)^2+1] \ge 0\\ \Leftrightarrow (x+1)(x-2)\ge 0(\text{Do } (x+1)^2+1>0 \ \forall \ x) (2)$

BXD:

\begin{array}{|c|cc|} \hline x&-\infty&&-1&&2&&+\infty\\\hline x+1&&-&0&+&|&+& \\\hline x-2&&-&|&-&0&+& \\\hline (x+1)(x-2) &&+&0&-&0&+&\\\hline \end{array}

Dựa vào BXD $\Rightarrow$ Nghiệm của $(2): \Leftrightarrow x\in (-\infty;1] \cup [2;+\infty).$