Question

Giải phương trình :

\(10.\left(\dfrac{x-2}{x+1}\right)^2+\left(\dfrac{x+2}{x-1}\right)^2=11.\dfrac{x^2-4}{x^2-1}\)

Answer 1

\(\Leftrightarrow10\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2=11\left(x^2-4\right)\left(x^2-1\right)\)

\(\Leftrightarrow10\left(x^4+9x^2+4-6x^3+4x^2-12x\right)+\left(x^4+9x^2+4+6x^3+4x^2+12x\right)=11\left(x^4-5x^2+4\right)\)

\(\Leftrightarrow10\left(x^4-6x^3+13x^2-12x+4\right)+x^4+6x^3+13x^2+12x+4-11x^4+55x^2-44=0\)

=>\(10x^4-60x^3+130x^2-120x+40-10x^4+6x^3+68x^2+12x-40=0\)

=>\(-54x^3+198x^2-108x=0\)

=>54x^3+198x^2+108x=0

=>9x(6x^2+33x+12)=0

=>\(x\in\left\{0;\dfrac{-11+\sqrt{89}}{4};\dfrac{-11-\sqrt{89}}{4}\right\}\)