E=(\(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)):(\(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-y}-\frac{-x+y}{\sqrt{xy}}\)
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Những câu hỏi liên quan
30 tháng 8 2020
Ta có: \(\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}-\frac{x+y}{\sqrt{xy}}\right)\)
\(=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\left(\frac{x\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}{\sqrt{xy}\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}-\sqrt{x}\right)}+\frac{y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}-\sqrt{x}\right)}-\frac{\left(x+y\right)\left(y-x\right)}{\sqrt{xy}\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}-\sqrt{x}\right)}\right)\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\left(\frac{x\sqrt{xy}-x^2+y\sqrt{xy}+y^2-\left(y^2-x^2\right)}{\sqrt{xy}\left(y-x\right)}\right)\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\left(\frac{x\sqrt{xy}+y\sqrt{xy}}{\sqrt{xy}\left(y-x\right)}\right)\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(y-x\right)}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x+y}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\frac{x+y}{\sqrt{y}+\sqrt{x}}\cdot\frac{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}-\sqrt{x}\right)}{x+y}\)
\(=\sqrt{y}-\sqrt{x}\)
11 tháng 8 2017
\(A=\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}-\frac{x+y}{\sqrt{xy}}\right)\)
\(=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\frac{x\left(\sqrt{xy}-x\right)\sqrt{xy}+y\left(\sqrt{xy}+y\right)\sqrt{xy}-\left(x+y\right)\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}{\sqrt{xy}\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x^2y-x^2\sqrt{xy}+xy^2+y^2\sqrt{xy}-y^2\sqrt{xy}+x^2\sqrt{xy}}{xy^2-x^2y}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy^2-x^2y}{xy^2+x^2y}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}{xy\left(x+y\right)}\)
\(=\sqrt{y}-\sqrt{x}\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 5 2019
ĐKXĐ:...
Để gõ công thức cho nhanh ta đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\)
\(\frac{a^2+b^2}{a+b}:\left(\frac{a^2+b^2}{ab}+\frac{b^2}{a^2-ab}-\frac{a^2}{b^2+ab}\right)=\frac{a^2+b^2}{ab}:\left(\frac{a^2+b^2}{ab}+\frac{b^2}{a\left(a-b\right)}-\frac{a^2}{b\left(a+b\right)}\right)\)
\(=\frac{a^2+b^2}{ab}:\left(\frac{\left(a^2+b^2\right)\left(a^2-b^2\right)+b^3\left(a+b\right)-a^3\left(a-b\right)}{ab\left(a-b\right)\left(a+b\right)}\right)\)
\(=\frac{a^2+b^2}{ab}:\left(\frac{a^4-b^4+ab^3+b^4-a^4+a^3b}{ab\left(a-b\right)\left(a+b\right)}\right)\)
\(=\frac{a^2+b^2}{ab}:\left(\frac{ab\left(a^2+b^2\right)}{ab\left(a-b\right)\left(a+b\right)}\right)=\frac{\left(a^2+b^2\right)\left(a-b\right)\left(a+b\right)}{a^2+b^2}=a^2-b^2=x-y\)
LT
2
BH
31 tháng 7 2019
\(=\left(\frac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\sqrt{x}}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{2\sqrt{xy}}\)
\(=\left(\frac{y-x}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{2\sqrt{xy}}\)
\(=\frac{\left(y-x\right)\left(\sqrt{x}-\sqrt{y}\right)}{2xy}\)
31 tháng 7 2019
\(\left(\frac{\sqrt{y}}{x+\sqrt{xy}}-\frac{\sqrt{y}}{x-\sqrt{xy}}\right).\frac{x-y}{2\sqrt{xy}}\)
\(=\frac{x-y}{2\sqrt{xy}}.\left[-\frac{2y\sqrt{x}}{\left(x+\sqrt{yx}\right)\left(x-\sqrt{yx}\right)}\right]\)
\(=-\frac{2y\sqrt{x}}{\left(x+\sqrt{xy}\right)\left(x-\sqrt{xy}\right)}.\frac{x-y}{2\sqrt{xy}}\)
\(=-\frac{2y\sqrt{x}.\left(x-y\right)}{\left(x+\sqrt{xy}\right)\left(x-\sqrt{xy}\right).2\sqrt{xy}}\)
\(=-\frac{y\sqrt{x}\left(x-y\right)}{\left(x+\sqrt{xy}\right)\left(x-\sqrt{xy}\right).\sqrt{xy}}\)
\(=-\frac{\sqrt{y}\left(x-y\right)}{\left(x+\sqrt{yx}\right)\left(x-\sqrt{yx}\right)}\)
\(=-\frac{\sqrt{y}}{x}\)