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a) điều kiện \(x;y\ge0\) ; \(x\ne y\)
b) ta có : \(A=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow A=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow A=\left(\sqrt{x}+\sqrt{y}\right)-\left(\sqrt{x}+\sqrt{y}\right)=0\)
\(a.ĐKXĐ:\left\{{}\begin{matrix}x\ne y\\x;y>0\end{matrix}\right.\)
\(b.A=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{x-2\sqrt{xy}+y+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}-\sqrt{x}-\sqrt{y}=\sqrt{x}+\sqrt{y}-\sqrt{x}-\sqrt{y}=0\)
a) ĐKXĐ: \(x,y\ge0\)
\(M=\dfrac{x\sqrt{y}-\sqrt{y}-y\sqrt{x}+\sqrt{x}}{1+\sqrt{xy}}=\dfrac{x\sqrt{y}-y\sqrt{x}+\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\)
\(=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\left(\sqrt{x}-\sqrt{y}\right)}{1+\sqrt{xy}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(1+\sqrt{xy}\right)}{1+\sqrt{xy}}=\sqrt{x}-\sqrt{y}\)
b) \(x=\left(1-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)
\(y=3-\sqrt{8}\Rightarrow\sqrt{y}=\sqrt{3-\sqrt{8}}=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)
\(\Rightarrow M=\left(\sqrt{3}-1\right)-\left(\sqrt{2}-1\right)=\sqrt{3}-\sqrt{2}\)
Ta có: \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt{xy}}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}\right)+\sqrt{y}\left(x-\sqrt{xy}\right)}{x^2-xy}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}+x-\sqrt{xy}\right)}{x\left(x-y\right)}=\dfrac{2x\sqrt{y}}{x\left(x-y\right)}\)
\(=\dfrac{2\sqrt{y}}{x-y}=\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}.\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1+1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)
Lời giải:
a) ĐK: $x\geq 0; y\geq 0; x\neq y$
\(A=\left[\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}-\frac{(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}+y)}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\right]:\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
b) \(1-A=\frac{(\sqrt{x}-\sqrt{y})^2}{x-\sqrt{xy}+y}>0\) với mọi $x\neq y; x,y\geq 0$
$\Rightarrow A< 1$
Lời giải:
\(A=\frac{x+y}{\sqrt{xy}}: \frac{x-y}{\sqrt{xy}}=\frac{x+y}{\sqrt{xy}}.\frac{\sqrt{xy}}{x-y}=\frac{x+y}{x-y}\)
\(=\frac{1+a+1-a}{1+a-(1-a)}=\frac{2}{2a}=\frac{1}{a}\)
a) tự làm.
b) \(P=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}-\dfrac{x\sqrt{x}-y\sqrt{y}}{x-y}\right)\div\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\left(\sqrt{x}+\sqrt{y}-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x+\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\dfrac{x+\sqrt{xy}+\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\sqrt{xy}\cdot\dfrac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}\)
\(=\dfrac{\sqrt{xy}}{x-2\sqrt{xy}+y+\sqrt{xy}}\)
\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)