Question

Помогите пожалуйста‍♀️

Answer 1

$1.\; \; (x+2)^2=(x-6)^2\\\\(x+2)^2-(x-6)^2=0\qquad [\; a^2-b^2=(a-b)(a+b)\; ]\\\\\Big ((x+2)-(x-6)\Big )\Big ((x+2)+(x-6)\Big )=0\\\\8\cdot (2x-4)=0\; \; \to \; \; 2x-4=0\; ,\; \; \underline {x=2}\\\\\\2.\; \; (7-x)^2=(x+3)^2\\\\(7-x-x-3)(7-x+x+3)=0\\\\(4-2x)\cdot 10=0\; \; \to \; \; 2x=4\; ,\; \underline {x=2}\\\\\\3.\; \; (x-1)^2=(14-x)^2\\\\(x-1-14+x)(x-1+14-x)=0\\\\(2x-15)\cdot 13=0\; \; \to \; \; \underline {x=7,5}\\\\\\4.\; \; \frac{6}{x+9}=-\frac{2}{3}\; \; \to \; \; 18=-2x-18\; ,\; \; 2x=-36\; ,\;\; \underline {x=-18}$

$5,\; \; \frac{10}{x-4}=\frac{5}{2}\; \; \to \; \; 20=5x-20\; ,\; \; 5x=40\; \; ,\; \; \underline {x=8}\\\\\\6.\; \; \frac{2}{x-6}=-5\; \; \to \; \; 2=-5x+30\; \; ,\; \; 5x=28\; \; ,\; \; \underline {x=5,6}\\\\\\7.\; \; 2x-4\geq 7x-1\; \; ,\; \; -3\geq 5x\; \; ,\; \; 5x\leq -3\; \; ,\; \; \underline {x\leq -0,6}\\\\\\8.\; \; 6x-2>4x-1\; \; ,\; \; 2x>1\; \; ,\; \; \underline {x>0,5}\\\\\\9.\; \; 3x-2(x-5)\leq -6\\\\3x-2x+10\leq -6\\\\\underline {x\leq-16}$