Question

  bài 2 ạ

Answer 1

Bài 2:

a: ĐKXĐ: x>0; x<>1

\(P=\left(1+\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{1}{x-\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-1+1}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}\left(\sqrt[]{x}-1\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)^2}=\dfrac{1}{\left(\sqrt{x}-1\right)^2}\)

b:, \(P\cdot\sqrt{5+2\sqrt{6}}\left(\sqrt{x}-1\right)^2=x-2005+\sqrt{2}+\sqrt{3}\)

=>\(\dfrac{1}{\left(\sqrt{x}-1\right)^2}\cdot\left(\sqrt{3}+\sqrt{2}\right)\cdot\left(\sqrt{x}-1\right)^2=x-2005+\sqrt{2}+\sqrt{3}\)

=>\(x-2005+\sqrt{2}+\sqrt{3}=\sqrt{3}+\sqrt{2}\)

=>x=2005(nhận)