Question

[Nguyễn Ý Nhi]

Đừng giải tóm tắt nghen :((

Answer 1

[Nguyễn Huy Tú]

a, \(A=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\frac{2}{1-x}\)ĐK : \(x\ne0;x\ge0\)

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

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

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

b, Thay \(x=\frac{9}{25}\Rightarrow\sqrt{x}=\frac{3}{5}\)vào biểu thức A ta được : 

\(\frac{\frac{3}{5}}{\frac{3}{5}+1}=\frac{\frac{3}{5}}{\frac{8}{5}}=\frac{3}{8}\)Vậy với x = 9/25 thì A = 3/8 

c, Để A nguyên khi : \(\sqrt{x}⋮\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}+1-1⋮\sqrt{x}+1\Leftrightarrow-1⋮\sqrt{x}+1\)

\(\Rightarrow\sqrt{x}+1\inƯ\left(-1\right)=\left\{\pm1\right\}\)

\(\sqrt{x}+1\)	1	-1
\(\sqrt{x}\)	0	-2
x	0	loại

Kết hợp với gt \(x\inℤ\)=> x = 0 ( tm ) thì A nguyên 

Answer 2

[♥̳H̳ả̳i̳̳Q̳̳u̳a̳̳y̳̳X̳̳e̳ ( •тєαм ν...]

a,ĐK:x≠4;x>0

b,A=($\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}$)*$\frac{\sqrt{x}-2}{\sqrt{x}}$

=$\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}$*$\frac{\sqrt{x}-2}{\sqrt{x}}$

=$\frac{\sqrt{x}-2}{\sqrt{x}}$

Để A>$\frac{1}{2}$thì$\frac{\sqrt{x}-2}{\sqrt{x}}$>$\frac{1}{2}$

⇔$\frac{\sqrt{x}-4}{2\sqrt{x}}$>0

⇔$\sqrt{x}-4>0\left(2\sqrt{x}>0\right)$

⇔ x>16(tm)

Để A>$\frac{1}{2}$thì 0<x>16và x≠4