a.

\(x=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\) \(\Rightarrow\sqrt{x}=\sqrt{5}+1\)

 \(\Rightarrow B=\dfrac{\sqrt{5}+1-1}{2+\sqrt{5}+1}=\dfrac{\sqrt{5}}{\sqrt{5}+3}=\dfrac{3\sqrt{5}-5}{4}\)

b.

\(B=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

B nguyên \(\Rightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Rightarrow\sqrt{x}+2=Ư\left(3\right)\)

Mà \(\sqrt{x}+2\ge2\Rightarrow\sqrt{x}+2=3\)

\(\Leftrightarrow\sqrt{x}=1\Rightarrow x=1\)

Mình cảm ơn.

a: \(x=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

\(\sqrt{x}-1\)

\(=\sqrt{\left(\sqrt5+1\right)^2}-1=\sqrt5+1-1=\sqrt5\)

\(2x+2-2x\cdot\sqrt{x}\)

\(=2\left(6+2\sqrt5\right)+2-2\cdot\left(6+2\sqrt5\right)\sqrt{\left(\sqrt5+1\right)^2}\)

\(=12+4\sqrt5+2-2\left(6+2\sqrt5\right)\left(\sqrt5+1\right)\)

\(=14+4\sqrt5-2\left(6\sqrt5+6+10+2\sqrt5\right)=14+4\sqrt5-2\left(8\sqrt5+16\right)\)

\(=14+4\sqrt5-16\sqrt5-32=-18-12\sqrt5\)

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

\(=\frac{\sqrt5}{-18-12\sqrt5}=\frac{-\sqrt5}{18+12\sqrt5}=\frac{-\sqrt5}{6\left(3+2\sqrt5\right)}=\frac{-\sqrt5\left(2\sqrt5-3\right)}{6\left(2\sqrt5+3\right)\left(2\sqrt5-3\right)}\)

\(=\frac{-10+3\sqrt5}{6\cdot\left(20-9\right)}=\frac{-10+3\sqrt5}{66}\)

bạn đặt \(\sqrt{x}=a\) , a> 0 

Thay \(\sqrt{x}=a\)  vô  biểu thức => rút gọn ra => thay trở lại  

giải chi tiết giúp mình đc không ạ?

a. \(C=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\frac{2\sqrt{x}-9-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

b. C=\(\frac{\sqrt{x}+1}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

C nguyên \(\Leftrightarrow\sqrt{x}-3\inƯ\left(4\right)\Rightarrow\sqrt{x}-3\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{1;2;4;5;7\right\}\Rightarrow x\in\left\{1;4;16;25;49\right\}\)

Vậy \(x\in\left\{1;4;16;25;49\right\}\)thì C nguyên