Question

giải giúp mình bài 2( đặc biệt là phần c) ạ , mình cảm ơn 

Answer 1

Bài 2:

a: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)

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

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

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

b: Để A<0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}< 0\)

=>\(\sqrt{x}-1< 0\)

=>\(\sqrt{x}< 1\)

=>0<=x<1

Kết hợp ĐKXĐ, ta được: 0<x<1

c: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1+2⋮\sqrt{x}-1\)

=>\(2⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1\in\left\{1;-1;2;-2\right\}\)

=>\(\sqrt{x}\in\left\{2;0;3;-1\right\}\)

=>\(\sqrt{x}\in\left\{0;2;3\right\}\)

=>\(x\in\left\{0;4;9\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{4;9\right\}\)