Question

Giải pt : \(\frac{2x^4}{\left(x^2+1\right)^2}+\frac{3x^2}{x^2+1}-2=0\)

Answer 1

\(\Leftrightarrow2x^4+3x^2\left(x^2+1\right)-2\left(x^2+1\right)^2=0\Leftrightarrow3x^4-x^2-2=0\)

\(x^2=1hoặcx^2=-\frac{2}{3}\left(L\right)\Leftrightarrow x=+-1\)

Answer 2

\(\frac{2x^4}{\left(x^2+1\right)^2}+\frac{3x^2}{x^2+1}-2=0\)

\(\frac{2x^2}{\left(x^2+1\right)^2}+\frac{3x^2\cdot\left(x^2+1\right)}{\left(x^2+1\right)^2}-2=0\)

\(2x^2+3x^4+3x^2-2=0\)

2*(x²-1)+3x²(x²+1)=0

đến đây thì chịu rồi