Question

x+1/x-2 + x-4/x+1 = 3x+3/x^2-x-2 помогите решить

Answer 1

$\frac{x+1}{x-2}+\frac{x-4}{x+1}=\frac{3x+3}{x^{2}-x-2}\\\\\frac{x+1}{x-2}+\frac{x-4}{x+1}-\frac{3x+3}{(x-2)(x+1)}=0\\\\\frac{(x+1)^{2}+(x-2)(x-4)-3x-3 }{(x-2)(x+1)}=0\\\\\frac{x^{2}+2x+1+x^{2}-4x-2x+8-3x-3}{(x-2)(x+1)}=0\\\\\frac{2x^{2}-7x+6 }{(x-2)(x+1)} =0$

$\left \{ {{2x^{2}-7x+6=0 } \atop {x-2\neq0;x+1\neq0}} \right.\\\\\left \{ {{2x^{2}-7x+6=0 } \atop {x\neq2,x\neq-1}} \right. \\\\2x^{2}-7x+6=0\\\\D=(-7)^{2} -4*2*6=49-48=1\\\\x_{1}=\frac{7-1}{4}=1,5\\\\x_{2}=\frac{7+1}{4}=2-neyd\\\\Otvet:\boxed{1,5}$