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Bài 2:
a: \(A=\frac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\frac{4-a}{\sqrt{a}-2}\)
\(=\frac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)=0\)
b: \(B=\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}:\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}\cdot\frac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2}=\frac{x-\sqrt{xy}+y}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
Bài 1:
a: \(A=\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b: \(B=\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\frac{\left(x-1\right)}{\sqrt{y}-1}\cdot\frac{\left|y-2\sqrt{y}+1\right|}{\left|\left(x-1\right)^2\right|}\)
\(=\left(x-1\right)\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\left(\sqrt{y}-1\right)}{x-1}\)
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 \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
Bài 3:
a: \(A=\frac{1}{2+\sqrt{x}}+\frac{1}{2-\sqrt{x}}+\frac{2\sqrt{x}}{4-x}\)
\(=\frac{1}{\sqrt{x}+2}-\frac{1}{\sqrt{x}-2}-\frac{2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-2-\sqrt{x}-2-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{-2\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=-\frac{2}{\sqrt{x}-2}\)
b: Thay x=3 vào A, ta được: \(A=-\frac{2}{\sqrt3-2}=\frac{2}{2-\sqrt3}=2\left(2+\sqrt3\right)=4+2\sqrt3\)
Bài 2:
a: \(A=\frac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\frac{4-a}{\sqrt{a}-2}\)
\(=\frac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)=0\)
b: \(B=\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}:\left(\sqrt{x}-\sqrt{y}\right)^2\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}\cdot\frac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2}=\frac{x-\sqrt{xy}+y}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
Bài 1:
a: \(A=\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
\(=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\left|\frac{\sqrt{x}-1}{\sqrt{x}+1}\right|=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b: \(B=\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}\)
\(=\frac{\left(x-1\right)}{\sqrt{y}-1}\cdot\frac{\left|y-2\sqrt{y}+1\right|}{\left|\left(x-1\right)^2\right|}\)
\(=\left(x-1\right)\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}-1\right)\left(x-1\right)^2}=\frac{\left(\sqrt{y}-1\right)}{x-1}\)