Question

Решите неравенство
4/(3х+1)^2 - 8/3x+1 + 3 ⩽ 0
Полное решение

Answer 1

$x \in [-\frac{1}{9}; \quad \frac{1}{3}]$

Объяснение:

ОДЗ:

$3x+1 \neq 0;$

$3x \neq -1;$

$x \neq -\frac{1}{3};$

$\frac{4}{(3x+1)^{2}}-\frac{8}{3x+1}+3 \leq 0 \quad | \quad \cdot (3x+1)^{2} \neq 0;$

$4-8(3x+1)+3(3x+1)^{2} \leq 0;$

$3(9x^{2}+6x+1)-24x-8+4 \leq 0;$

$27x^{2}+18x+3-24x-4 \leq 0;$

$27x^{2}-6x-1 \leq 0;$

Найдём нули функции:

$27x^{2}-6x-1=0;$

$D=b^{2}-4ac;$

$D=(-6)^{2}-4 \cdot 27 \cdot (-1)=36+108=144;$

$x_{1,2}=\frac{-b \pm \sqrt{D}}{2a};$

$x_{1}=\frac{-(-6)+\sqrt{144}}{2 \cdot 27}=\frac{6+12}{54}=\frac{18}{54}=\frac{1}{3};$

$x_{2}=\frac{-(-6)-\sqrt{144}}{2 \cdot 27}=\frac{6-12}{54}=\frac{-6}{54}=-\frac{1}{9};$

$x=1 \Rightarrow 27 \cdot 1^{2}-6 \cdot 1-1=27-6-1=200;$

$x=0 \Rightarrow 27 \cdot 0^{2}-6 \cdot 0-1=0-0-1=-1$

$-\frac{1}{3}=-\frac{3}{9};$

$x=-\frac{2}{9} \Rightarrow 27 \cdot (-\frac{2}{9})^{2}-6 \cdot (-\frac{2}{9})-1=27 \cdot \frac{4}{81}+6 \cdot \frac{2}{9}-1=\frac{4}{3}+\frac{4}{3}-1=\frac{5}{3}0;$

$x=-1 \Rightarrow 27 \cdot (-1)^{2}-6 \cdot (-1)-1=27 \cdot 1 +6 \cdot 1-1=27+6-1=320;$

$x \in [-\frac{1}{9}; \quad \frac{1}{3}];$