a)\(\hept{\begin{cases}a\ge0\\\sqrt{a}-2>0\Leftrightarrow\\\sqrt{a}+2>0\end{cases}a>4}\)

b)\(\frac{\sqrt{a}\left(\sqrt{a}+2\right)+\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}.\frac{a-4}{2\sqrt{a}}\)  \(=\frac{2a}{a-4}.\frac{a-4}{2\sqrt{a}}=\sqrt{a}\)

c)\(\sqrt{a}>3\Leftrightarrow a>9\)

dsdsdsdsd

dsdsdsdsd

dsdsdssd

dsdsdssds

\(\left(a\right)\hept{\begin{cases}a\ge0\\\sqrt{a}-2\ne0\Rightarrow a\ne4\\\sqrt{a}+2\ne0\end{cases}}\)\(\hept{\begin{cases}a\ge0\\a\ne4\end{cases}}\) OK nếu bạn hỏi thật bạn nhận dduocj sự giúp đỡ chân thật là vậy"

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

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

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

\(A=2\cdot\frac{a-2}{a+2}\left(a\ne0;a\ne\pm2\right)\)

b) để A = 1 => \(2\cdot\frac{a-2}{a+2}=1\)

=> 2a - 4 = a + 2

=> a = 6 (thỏa mãn)

bài 2) a) ĐKXĐ: \(x\ne4\)

b) \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)

\(\Leftrightarrow B=\frac{2\sqrt{x}+\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(\Rightarrow B=\frac{2\sqrt{x}+4}{x-4}=\frac{2}{\sqrt{x}-2}\)

c) \(B=\frac{2}{\sqrt{3+2\sqrt{3}}-2}\) \(\approx3,69\)

(bạn tự bấm máy tính nhé nhưng theo mình thấy nếu x = 4 + 2\(\sqrt{3}\) hay \(3+2\sqrt{2}\) thì sẽ cho kết quả đẹp hơn, k biết bạn có nhầm đề k nữa!)

bài 3)

a, \(B=\left(\frac{1}{\sqrt{y}+1}-\frac{3\sqrt{y}}{\sqrt{y}-1}+3\right)\cdot\frac{\sqrt{y}+1}{\sqrt{y}+2}\left(y\ne1;y\ne4\right)\)

\(\Leftrightarrow B=\frac{\sqrt{y}-1-3y-3\sqrt{y}+3y-3}{y-1}\cdot\frac{\sqrt{y}+1}{\sqrt{y}+2}\)

\(\Rightarrow B=\frac{-2\sqrt{y}-4}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\cdot\frac{\sqrt{y}+1}{\sqrt{y}+2}\Rightarrow B=\frac{-2}{\sqrt{y}-1}\)

b) y = \(3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\)

=> B = \(\frac{-2}{\sqrt{\left(1+\sqrt{2}\right)^2}-1}\)

\(\Rightarrow B=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)

Ai giải giúp mình bài 1 với bài 4 trước đi

1,

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

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

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

\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)

Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)

2, 

a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)

b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)

\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)

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

c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)

Lời giải:

a) ĐK: $a>0; a\neq 1$

b)

\(B=\left(\frac{\sqrt{a}+2}{(\sqrt{a}+1)^2}-\frac{\sqrt{a}-2}{a-1}\right).\frac{\sqrt{a}+1}{\sqrt{a}}=\frac{\sqrt{a}+2}{(\sqrt{a}+1)^2}.\frac{\sqrt{a}+1}{\sqrt{a}}-\frac{\sqrt{a}-2}{a-1}.\frac{\sqrt{a}+1}{\sqrt{a}}\)

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

\(=\frac{(a+\sqrt{a}-2)-(a-\sqrt{a}-2)}{(a-1)\sqrt{a}}=\frac{2\sqrt{a}}{\sqrt{a}(a-1)}=\frac{2}{a-1}\) (đpcm)

Bài 2: 

\(\Leftrightarrow3\sqrt{x+5}-2\sqrt{x+5}=7\)

\(\Leftrightarrow\sqrt{x+5}=7\)

=>x+5=25

hay x=18