Question

Giải phương trình:

a, \(\left(x+1+\dfrac{1}{x}\right)^2=\left(x-1-\dfrac{1}{x}\right)^2\)
b,  \(\left(x-1\right)^2+3x^2=0\)

Answer 1

a) \(\left(x+1+\dfrac{1}{x}\right)^2=\left(x-1-\dfrac{1}{x}\right)^2\\ \Leftrightarrow\left(x+1+\dfrac{1}{x}\right)^2-\left(x-1-\dfrac{1}{x}\right)^2=0\\ \Leftrightarrow\left(x+1+\dfrac{1}{x}-x+1+\dfrac{1}{x}\right)\left(x+1+\dfrac{1}{x}+x-1-\dfrac{1}{x}\right)=0\\ \Leftrightarrow2\left(1+\dfrac{1}{x}\right)\cdot2x=0\\ \Leftrightarrow4x\left(1+\dfrac{1}{x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(S=\left\{-1;0\right\}\) là tập nghiệm của pt.

Answer 2

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

\(\Leftrightarrow x^2-2x+1+3x^2=0\)

\(\Leftrightarrow4x^2-2x+1=0\)

\(\text{Δ}=\left(-2\right)^2-4\cdot4\cdot1=4-16=-12< 0\)

=> Phương trình vô nghiệm

Vậy: \(S=\varnothing\)