Question

ПОЖАЛУЙСТА ПОМОГИТЕ РЕШИТЬ ДВА ПРИМЕРА!

1. (27*3)^-4 * (9^-1)^-2

2. x-1/x+2 + x+1/x-2 + 2x+8/4-x^2 = 0

Answer 1

1)

$(27*3)^-^4*(9^-^1)^-^2=27^-^4*3^-^4*(\frac{1}{9})^-^2=(3^3)^-^4*3^-^4*9^2=\\=3^-^1^2*3^-^4*(3^2)^2=3^-^1^6*3^4=3^-^1^2$

Ответ: 3⁻¹².

2)

$\frac{x-1}{x+2}+\frac{x+1}{x-2}+\frac{2x+8}{4-x^2}=0\\\\\frac{x-1}{x+2}-\frac{x+1}{2-x}+\frac{2x+8}{(2-x)(2+x)}=0\\\\\frac{(x-1)(2-x)-(x+1)(2+x)+(2x+8)}{(2-x)(2+x)}=0\\\\\frac{-2x^2+2x+4}{(2-x)(2+x)}=0\\\\OD3:\ x\neq2\ and\ x\neq-2\\\\-2x^2+2x+4=0\\2x^2-2x-4=0\\x^2-x-2=0\\D=(-1)^2-4*1*(-2)=9\\\sqrt{9}=3\\\\x_1=\frac{1-3}{2}=-1\\\\x_2=\frac{1+3}{2}=2$

Ответ: -1.