Question

Найдите корни уравнений: 3x+1/x-2=2x-10/x+1 ; x+2/x-1+x/x+1=6/x^2-1

Answer 1

$\displaystyle 1.\quad \frac{3x+1}{x-2} =\frac{2x-10}{x+1} \\\\\\(3x+1)(x+1)=(2x-10)(x-2)\\\\3x^2+3x+x+1=2x^2-4x-10x+20\\\\x^2+18x-19=0\\\\a=1;\quad b=18;\quad c=-19\\\\a+b+c=0 \quad \Rightarrow \quad x_1=1;\quad x_2=\frac{c}{a} \\\\x_1=1;\quad x_2=-19\\\\Otvet:\quad -19;\, 1.\\\\\\2.\quad \frac{x+2}{x-1}+\frac{x}{x+1} =\frac{6}{x^2-1} \\\\\\\frac{(x+2)(x+1)}{(x-1)(x+1)} +\frac{x(x-1)}{(x-1)(x+1)}=\frac{6}{(x-1)(x+1)}\\\\\\(x+2)(x+1)+x(x-1)=6\quad\quad ,\, x\neq \pm1\\\\x^2+x+2x+2+x^2-x-6=0\\\\ 2x^2+2x-4=0$

$\displaystyle a=2;\quad b=2;\quad c=-4\\\\a+b+c=0 \quad \Rightarrow \quad x_1=1;\quad x_2=\frac{c}{a} \\\\x_1\neq 1;\quad x_2=-2\\\\Otvet:\quad -2.$