Bài 3:
1: \(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x}-2}\)
2: Thay x=25 vào A, ta được:
\(A=\frac{\sqrt{25}}{\sqrt{25}-2}=\frac{5}{5-2}=\frac53\)
3: \(A=-\frac13\)
=>\(\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{-1}{3}\)
=>\(3\sqrt{x}=-\sqrt{x}+2\)
=>\(4\sqrt{x}=2\)
=>\(\sqrt{x}=\frac24=\frac12\)
=>x=1/4(nhận)
Bài 2:
a: \(A=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\frac{x-1}{2\sqrt{x}}\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2+\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\sqrt{x}\left(x-2\sqrt{x}+1+x+2\sqrt{x}+1\right)}{2\sqrt{x}}=\frac{2x+2}{2}=x+1\)
b: A>-6
=>x+1>-6
=>x>-7
Kết hợp ĐKXĐ, ta được: x>0 và x<>1
