Ta có:
\(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow x^4+2x^2+1+3x^3+3x+2x^2=0\)
\(\Leftrightarrow x^4+3x^3+4x^2+3x+1=0\)
Xét x = 0 không là nghiệm của pt
Chia 2 vế của pt cho x2 ta được:
\(x^2+3x+4+\dfrac{3}{x}+\dfrac{1}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}\right)+\left(3x+\dfrac{3}{x}\right)+4=0\)
\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}\right)+3\left(x+\dfrac{1}{x}\right)+4=0\)
Đặt a = \(x+\dfrac{1}{x}\)
\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}=a^2\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}=a^2-2\)
Suy ra:
\(\Leftrightarrow\left(a^2-2\right)+3a+4=0\)
\(\Leftrightarrow a^2+3a+2=0\)
\(\Leftrightarrow\left(a+1\right)\left(a+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-1\\a=-2\end{matrix}\right.\)
Với a = -1
=> \(x+\dfrac{1}{x}=-1\)
\(\Rightarrow x^2+1=-x\) (loại)
Với a = -2
=> \(x+\dfrac{1}{x}=-2\)
\(\Rightarrow x^2+1+2x=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x=-1\)
Vây pt có tập nghiệm là: \(S=\left\{-1\right\}\)
Ta có:\(2a^2+2b^2=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow2a^2-ab-4ab+2b^2=0\)
\(\Leftrightarrow a\left(2a-b\right)-2b\left(2a-b\right)=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-2b=0\left(loai\right)\\2a-b=0\end{matrix}\right.\Leftrightarrow2a=b\)
\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{a+2a}{a-2a}=-3\)
1)Cách khác:
\(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)^2+x\left(x^2+1\right)+2x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+x+1\right)+2x\left(x^2+1+x\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=0\left(loai\right)\\x^2+2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)