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\(A=\left(\dfrac{4\sqrt{xy}+x-2\sqrt{xy}+y}{2\left(x-y\right)}\right)\cdot\dfrac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\left(x-y\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=1\)
Bài 1:
a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)
Do đó: A>=0
\(B=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{xy}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\left(\sqrt{x^3}+x\right)}=\dfrac{\left(\sqrt{x}+1\right)\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)x\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}}\)
a: \(N=\dfrac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)-x\sqrt{x}+y\sqrt{y}}{x-y}\right)\)
\(=\dfrac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}:\dfrac{x\sqrt{x}+x\sqrt{y}-y\sqrt{x}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}{x-y}\)
\(=\dfrac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{x-y}{x\sqrt{y}-y\sqrt{x}}\)
\(=\dfrac{x-\sqrt{xy}+y}{1}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}\)
\(=\dfrac{x-\sqrt{xy}+y}{\sqrt{xy}}\)
b: \(N-1=\dfrac{x-2\sqrt{xy}+y}{\sqrt{xy}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{xy}}>0\)
=>N>1
Ta có \(A=\left(\frac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\)
\(=\left(\frac{4\sqrt{xy}+\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\) (Quy đồng biểu thức đầu và đổi dấu số hạng cuối)
\(=\left(\frac{4\sqrt{xy}+x-2\sqrt{xy}+y}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=1.\)
Vậy giá trị biểu thức \(A=1.\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
\(A=\left(5-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\right)\left(5+\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\right)\)
\(=\left(5-\sqrt{xy}\right)\left(5+\sqrt{xy}\right)\)
=5-xy