Question

допоможіть 8) (x+4)(x-2)-(x+5)(x+3) ≤ -8.x; 9) (3x + 1)²-(x + 2)(4x-1) > 5(x-1)² +7x; 10) 3x(5+12x)-(6x-1)(6x +1) ≥10x.​

Answer 1

Восьмое задание

$\left(x-2\right)\,\left(x+4\right)-\left(x+3\right)\,\left(x+5\right)\leq -8\,x\\x^2+4x-2x-8-\left ( x^2+5x+3x+15 \right )\leq -8x\\x^2+2x-8-x^2-8x-15\leq -8x\\-6x-23\leq -8x\Leftrightarrow 2x\leq 23\Rightarrow x\leq \frac{23}{2}$

Девятое задание

$\left(3\,x+1\right)^{2}-\left(x+2\right)\,\left(4\,x-1\right) > 7\,x+5\,\left(x-1\right)^{2}\\9x^2+6x+1-\left ( 4x^2-x+8x-2 \right ) > 7x+5\left ( x^2-2x+1 \right )\\9x^2+6x+1-4x^2-7x+2 > 7x+5x^2-10x+5\\9x^2-4x^2-5x^2+6x-7x-7x+10x+1+2-5 > 0\\2x-2 > 0\Rightarrow x > 1$

Десятое задание

$3\,x\,\left(12\,x+5\right)-\left(6\,x-1\right)\,\left(6\,x+1\right)\geq 10\,x\\36x^2+15x-\left ( 36x^2-1 \right )-10x\geq 0\\36x^2+15x-36x^2+1-10x\geq 0\\5x+1\geq 0\Rightarrow x\geq -\frac{1}{5}$

Answer 2

$8)(x+4)(x-2)-(x+5)(x+3)\leq -8x\\x^2-2x+4x-8-(x^2+3x+5x+15)\leq -8x\\x^2-2x+4x-8-x^2-3x-5x-15\leq -8x\\x^2-x^2-2x+4x-3x-5x+8x\leq 8+15\\2x\leq 23\\x\leq 11.5\\x\in(-\infty;11.5]$

$9)(3x+1)^2-(x+2)(4x-1) > 5(x-1)^2+7x\\9x^2+6x+1-(4x^2-x+8x-2) > 5(x^2-2x+1)+7x\\9x^2+6x+1-4x^2+x-8x+2 > 5x^2-10x+5+7x\\9x^2-4x^2-5x^2+6x+x-8x+10x-7x > 5-2-1\\2x > 2\\x > 1\\x\in(1;+\infty)$

$10)3x(5+12x)-(6x-1)(6x+1)\geq 10x\\15x+36x^2-36x^2+1\geq 10x\\15x-10x\geq -1\\5x\geq -1\\x\geq -0.2\\x\in[-0.2;+\infty)$