Question

Giải phương trình sau:

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

 

Answer 1

Sửa đề: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

Ta có: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

\(\Leftrightarrow\left(x^2+1\right)^2+2x\left(x^2+1\right)+x\left(x^2+1\right)+2x^2=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2+2x+1\right)+x\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)\left(x^2+x+1\right)=0\)

mà \(x^2+x+1>0\forall x\)

nên \(x^2+2x+1=0\)

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

Vậy: S={-1}