Question

2.183. Решите уравнение дам 20б

Answer 1

РЕШЕНИЕ СМОТРИ НА ФОТОГРАФИИ

Объяснение:

Answer 2

$\frac{30}{x^{2}-1 }+\frac{7-18x}{x^{3}+1 }=\frac{13}{x^{2}-x+1 } \\\\\frac{30}{(x+1)(x-1) }+\frac{7-18x}{(x+1)(x^{2}-x+1)}-\frac{13}{x^{2}-x+1 }=0\\\\\frac{30(x^{2}-x+1)+(7-18x)(x-1)-13(x^{2}-1)}{(x+1)(x-1)(x^{2}-x+1) }=0\\\\\frac{30x^{2}-30x+30+7x-7-18x^{2} +18x-13x^{2}+13}{(x+1)(x-1)(x^{2}-x+1 }=0\\\\\frac{-x^{2}-5x+36 }{(x+1)(x-1)(x^{2}-x+1) }=0\\\\\frac{x^{2}+5x-36 }{(x+1)(x-1)(x^{2}-x+1) }=0\\\\\frac{(x-4)(x+9)}{(x-1)(x^{3}+1) }=0\\\\x\neq-1 \ ; \ x\neq 1$

$\left[\begin{array}{ccc}x-4=0\\x+9=0\end{array}\right\\\\\left[\begin{array}{ccc}x_{1}=4 \\x_{2}=-9 \end{array}\right\\\\Otvet:\boxed{4 \ ; \ - 9}$