Question

(x+1/x)^2-3(x+1/x)+2=0

Answer 1

ĐKXĐ: x<>0

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

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

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

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

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

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

mà \(x^2-x+1=\left(x-\frac12\right)^2+\frac34\ge\frac34>0\forall x\) thỏa mãn ĐKXĐ

nên \(x^2-2x+1=0\)

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

=>x-1=0

=>x=1