Question

Giải PT này giùm mk nha : \(10\cdot\left(\dfrac{x-2}{x+1}\right)^2+\left(\dfrac{x+2}{x-1}\right)^2-11\cdot\left(\dfrac{x^2-4}{x^2-1}\right)=0\)

Answer 1

\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2}{\left(x^2-1\right)^2}-\dfrac{11\left(x^4-5x^2+4\right)}{\left(x^2-1\right)^2}=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)^2-6x\left(x^2+2\right)+9x^2+\left(x^2+2\right)^2+6x\left(x^2+2\right)+9x^2-11\left(x^4-5x^2+4\right)=0\)

\(\Leftrightarrow2\left(x^2+2\right)^2+18x^2-11x^4+55x^2-44=0\)

\(\Leftrightarrow2\left(x^4+4x^2+4\right)-11x^4+73x^2-44=0\)

=>\(-9x^4+81x^2-36=0\)

=>9x^4-81x^2+36=0

=>x^4-9x^2+4=0

=>\(x^2=\dfrac{9\pm\sqrt{65}}{2}\)

=>\(x=\pm\sqrt{\dfrac{9\pm\sqrt{65}}{2}}\)