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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 \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
a) \(A=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\left(a>0;a\ne1\right)\)
\(=\left[\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right]:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)
b) Để \(A=\frac{1}{2}\)
\(\Leftrightarrow\frac{\sqrt{a}-1}{\sqrt{a}}=\frac{1}{2}\)
\(\Leftrightarrow2\sqrt{a}-2=\sqrt{a}\)
\(\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\left(tm\right)\)
Mk làm từng câu nhé !
a)\(A=\frac{x-\sqrt{x}}{x-1}\left(đk:x\ge0,x\ne1\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)(vì \(x\ge0\))
\(=\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(B=\frac{x-4}{x+2\sqrt{x}}\left(đk:x>0,x\ne4\right)\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=1+\frac{2}{\sqrt{x}}\)
a.\(DK:x\ge0,x\ne1\)
\(A=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(DK:x\ge0\)
\(B=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}}\)
b.\(A-B=\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{x-x+\sqrt{x}+2}{x+\sqrt{x}}=\frac{\sqrt{x}+2}{x+\sqrt{x}}>0\)
\(\Rightarrow A-B>0\Rightarrow A>B\)
c.Ta co:\(A.B=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
De \(A.B\in Z\)
\(\Rightarrow1-\frac{3}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{3}{\sqrt{x}+1}\in Z\)
\(\Rightarrow3⋮\sqrt{x}+1\)
\(\Rightarrow x=4\)
d.Ta co: \(A.B=\frac{\sqrt{x}-2}{\sqrt{x}+1}< \frac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}-4< \sqrt{x}+1\)
\(\Leftrightarrow x< 25\)
a) A=\(\left(\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\right)\times\frac{x-4}{\sqrt{x}+3}\) (dkxd:x # 4 ,x# 9)
=> A=\(\frac{\sqrt{x}+2-\sqrt{x}+2}{x-4}\times\frac{x-4}{\sqrt{x}+3}\)
=> A=\(\frac{4}{\sqrt{x}+3}\)
b) A>1/2 <=> \(\frac{4}{\sqrt{x}+3}>\frac{1}{2}\Leftrightarrow\sqrt{x}+3< 8\Leftrightarrow\sqrt{x}< 5\Leftrightarrow x< 25\) (tmdkxd)
Vay .....
bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)