Question

Giúp mình câu này với ạ

Answer 1

ĐKXĐ: x>=2

Ta có: \(x-3\sqrt{x-2}=\sqrt{x^2-2x}-3\sqrt{x}+2\)

=>\(x-11-3\sqrt{x-2}+9=\sqrt{x^2-2x}-\sqrt{9x}\)

=>\(x-11-3\left(\sqrt{x-2}-3\right)=\frac{x^2-2x-9x}{\sqrt{x^2-2x}+3\sqrt{x}}\)

=>\(x-11-3\cdot\frac{x-2-9}{\sqrt{x-2}+3}=\frac{x\left(x-11\right)}{\sqrt{x^2-2x}+3\sqrt{x}}\)

=>\(x-11-3\cdot\frac{x-11}{\sqrt{x-2}+3}=\frac{x\left(x-11\right)}{\sqrt{x^2-2x}+3\sqrt{x}}\)

=>\(\left(x-11\right)\left(1-\frac{3}{\sqrt{x-2}+3}\right)=\left(x-11\right)\cdot\frac{x}{\sqrt{x^2-2x}+3\sqrt{x}}\)

=>\(\left(x-11\right)\left(1-\frac{3}{\sqrt{x-2}+3}-\frac{x}{\sqrt{x^2-2x}+3\sqrt{x}}\right)=0\)

=>x-11=0

=>x=11(nhận)