Question

[kiều thanh tùng]

giúp với mọi người 

Answer 1

[Đinh Việt Sơn]

chịu thôi

Answer 2

[Nguyễn Huy Tú]

Bài 1 : Với \(x>0;x\ne1\)

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

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

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

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

\(=\frac{2\sqrt{x}}{\sqrt{x}}:\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{2\sqrt{x}+2}{\sqrt{x}-1}\)

b, Ta có : \(\frac{2\sqrt{x}+2}{\sqrt{x}-1}=\frac{2\left(\sqrt{x}-1\right)+5}{\sqrt{x}-1}=2+\frac{5}{\sqrt{x}-1}\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\sqrt{x}-1\)	1	-1	5	-5
x	4	0	36	loại