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AH
Akai Haruma
Giáo viên
29 tháng 12 2018
Lời giải:
a) ĐK: \(x>0; y> 0\)
\(P=\frac{(\sqrt{x}-\sqrt{y})^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)
\(=\frac{x-2\sqrt{xy}+y+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{xy}(\sqrt{x}-\sqrt{y})}{\sqrt{xy}}\)
\(=\frac{x+2\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}-(\sqrt{x}-\sqrt{y})\)
\(=\frac{(\sqrt{x}+\sqrt{y})^2}{\sqrt{x}+\sqrt{y}}-(\sqrt{x}-\sqrt{y})=(\sqrt{x}+\sqrt{y})-(\sqrt{x}-\sqrt{y})=2\sqrt{y}\)
b)
Khi \(y=4-2\sqrt{3}=3+1-2\sqrt{3.1}=(\sqrt{3}-1)^2\)
\(\Rightarrow \sqrt{y}=\sqrt{3}-1\)
\(\Rightarrow P=2\sqrt{y}=2(\sqrt{3}-1)\)
AH
Akai Haruma
Giáo viên
24 tháng 8 2018
Lời giải:
ĐK: \(x\geq 0; x\neq 1; x\neq 4y; y>0\)
\(B=\frac{\sqrt{x^3}}{\sqrt{y}(\sqrt{x}-2\sqrt{y})}-\frac{2x}{(x-2\sqrt{xy})+(\sqrt{x}-2\sqrt{y})}.\frac{(1-\sqrt{x})(1+\sqrt{x})}{1-\sqrt{x}}\)
\(=\frac{\sqrt{x^3}}{\sqrt{y}(\sqrt{x}-2\sqrt{y})}-\frac{2x}{(\sqrt{x}-2\sqrt{y})(\sqrt{x}+1)}.(1+\sqrt{x})\)
\(=\frac{\sqrt{x^3}}{\sqrt{y}(\sqrt{x}-2\sqrt{y})}-\frac{2x}{\sqrt{x}-2\sqrt{y}}\)
\(=\frac{\sqrt{x^3}-2x\sqrt{y}}{\sqrt{y}(\sqrt{x}-2\sqrt{y})}\)
\(=\frac{x(\sqrt{x}-2\sqrt{y})}{\sqrt{y}(\sqrt{x}-2\sqrt{y})}=\frac{x}{\sqrt{y}}\)
21 tháng 5 2022
a: \(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y}{x-y}\cdot\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\dfrac{4xy}{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
b: \(=\sqrt{x}+\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)-2\sqrt{y}\)
\(=\sqrt{x}-\sqrt{y}-\sqrt{x}+\sqrt{y}=0\)
c: \(=\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}\cdot\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
30 tháng 4 2021
Ta có: \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt{xy}}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}\right)+\sqrt{y}\left(x-\sqrt{xy}\right)}{x^2-xy}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}+x-\sqrt{xy}\right)}{x\left(x-y\right)}=\dfrac{2x\sqrt{y}}{x\left(x-y\right)}\)
\(=\dfrac{2\sqrt{y}}{x-y}=\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}.\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1+1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)
29 tháng 8 2021
Bài 2:
Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
24 tháng 9 2022
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
ĐKXĐ: \(\sqrt{y}\left(\sqrt{x}-2\right)< >0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y>0\\x\in[0;+\infty)\backslash\left\{4\right\}\end{matrix}\right.\)