Question

câu 4 nha

Answer 1

4:

ĐKXĐ: x<>0

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

\(2x - \frac{1}{x} + 1 = 2\left(x - \frac{1}{2x}\right) + 1\)

Do đó: Phương trình ban đầu sẽ trở thành: \(\left(x-\frac{1}{2x}\right)^2+1=2\left(x-\frac{1}{2x}\right)+1\)

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

=>\(\left(x-\frac{1}{2x}\right)\left(x-\frac{1}{2x}-2\right)=0\)

TH1: \(x-\frac{1}{2x}=0\)

=>\(2x^2-1=0\)

=>\(x^2=\frac12=\frac24\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{2}\left(nhận\right)\\ x=-\frac{\sqrt2}{2}\left(nhận\right)\end{array}\right.\)

TH2: \(x-\frac{1}{2x}-2=0\)

=>\(\frac{2x^2-1-4x}{2x}=0\)

=>\(2x^2-4x-1=0\)

=>\(x^2-2x-\frac12=0\)

=>\(x^2-2x+1-\frac32=0\)

=>\(\left(x-1\right)^2=\frac32=\frac64\)

=>\(\left[\begin{array}{l}x-1=\frac{\sqrt6}{2}\\ x-1=-\frac{\sqrt6}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt6+2}{2}\left(nhận\right)\\ x=\frac{-\sqrt6+2}{2}\left(nhận\right)\end{array}\right.\)