Решите неравенство ..................................................................

Question 1

Решите неравенство
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Question 2

$\frac{4x+12}{x^2+x-2} \geq \frac{3x+9}{5+3x-2x^2}$

$\frac{4x+12}{x^2+x-2} -\frac{3x+9}{5+3x-2x^2} \geq 0$

$\frac{4x+12}{x^2+x-2} -\frac{3x+9}{-(2x^2-3x-5)} \geq 0$

$\frac{4x+12}{x^2+x-2}+\frac{3x+9}{2x^2-3x-5} \geq 0$

$x^2+x-2=0$

$D=1^2-4*1*(-2)=1+8=9$

$x_1= \frac{-1+3}{2}=1$

$x_2= \frac{-1-3}{2}=-2$

$x^2+x-2=(x-1)(x+2)$

$2x^2-3x-5=0$

$D=(-3)^2-4*2*(-5)=9+40=49$

$x_1= \frac{3+7}{4}=2.5$

$x_2= \frac{3-7}{4}=-1$

$2x^2-3x-5=2(x-2.5)(x+1)$

$\frac{4(x+3)}{(x-1)(x+2)}+\frac{3(x+3)}{2(x-2.5)(x+1)} \geq 0$

$\frac{4(x+3)(2x^2-3x-5)+3(x+3)(x^2+x-2)}{(x-1)(x+2)(2x-5)(x+1)} \geq 0$

$\frac{(x+3)(4(2x^2-3x-5)+3(x^2+x-2))}{(x-1)(x+2)(2x-5)(x+1)} \geq 0$

$\frac{(x+3)(8x^2-12x-20+3x^2+3x-6)}{(x-1)(x+2)(2x-5)(x+1)} \geq 0$

$\frac{(x+3)(11x^2-9x-26)}{(x-1)(x+2)(2x-5)(x+1)} \geq 0$

$11x^2-9x-26=0$

$D=(-9)^2-4*11*(-26)=81+1144=1225=35^2$

$x_1= \frac{9+35}{22}=2$

$x_2= \frac{9-35}{22}=- \frac{13}{11} =-1 \frac{2}{11}$

$11x^2-9x-26=11(x-2)(x+1 \frac{2}{11} )$

$\frac{(x+3)*11(x-2)(x+1 \frac{2}{11} )}{(x-1)(x+2)(2x-5)(x+1)} \geq 0$

$\frac{(x+3)(x-2)(11x+13 )}{(x-1)(x+2)(2x-5)(x+1)} \geq 0$

- + - + - + - +
--------[-3]--------(-2)--------[-1 2/11]-------(-1)-------(1)--------[2]--------(2.5)------
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$x$ ∈ $[-3;-2)$ ∪ $[-1 \frac{2}{11} ;-1)$ ∪ $(1;2]$ ∪ $(2.5;+$ ∞ $)$

Question 3

Спасибо