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\(A=\left\{\frac{2\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{x}\left(x+y\right)}{\sqrt{x}}\right\}.\left(\frac{\sqrt{x}-\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\right)^2.\)
=> \(A=\left(2\sqrt{xy}+x+y\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)
=> \(A=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)^2}=1\)
ĐS: A=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\left(\sqrt{x}+\sqrt{y}\right)}{x\sqrt{x}+y\sqrt{y}}\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{x\sqrt{x}+y\sqrt{y}}{x-y}\right)\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}^3+\sqrt{y}^3}\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x}^3+\sqrt{y}^3}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\right)\)
\(=\frac{1}{x-\sqrt{xy}+y}\left(\sqrt{x}+\sqrt{y}-\frac{x-\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}\right)\)
\(=\frac{1}{x-\sqrt{xy}+y}\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{x-\sqrt{xy}+y}{\sqrt{x}-\sqrt{y}}\right)\)
\(=\frac{1}{x-\sqrt{xy}+y}\left(\frac{x-y-x+\sqrt{xy}-y}{\sqrt{x}-\sqrt{y}}\right)\)
\(=\frac{1}{x-\sqrt{xy}+y}\left(\frac{\sqrt{xy}-2y}{\sqrt{x}-\sqrt{y}}\right)\)
tự làm tiếp nh đến đây dễ rồi
Năm 1930 có sự kiện gì và năm 1945 có sự kiện gì toán lóp 4
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\(A=\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}-\frac{x+y}{\sqrt{xy}}\right)\)
\(=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}:\frac{x\left(\sqrt{xy}-x\right)\sqrt{xy}+y\left(\sqrt{xy}+y\right)\sqrt{xy}-\left(x+y\right)\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}{\sqrt{xy}\left(\sqrt{xy}+y\right)\left(\sqrt{xy}-x\right)}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x^2y-x^2\sqrt{xy}+xy^2+y^2\sqrt{xy}-y^2\sqrt{xy}+x^2\sqrt{xy}}{xy^2-x^2y}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy^2-x^2y}{xy^2+x^2y}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{xy\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}{xy\left(x+y\right)}\)
\(=\sqrt{y}-\sqrt{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(5-\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\right)\)
\(A=\left[5-\frac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\right]\left[5+\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\right]\)
\(A=\left[5-\sqrt{xy}\right]\left[5+\sqrt{xy}\right]\)
\(A=25-xy\)
vậy \(A=25-xy\)
\(A=\left(5-\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\right)\)
\(A=\left(5-\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\right)\left(5+\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\right)\)
\(A=\left(5-\sqrt{xy}\right)\left(5+\sqrt{xy}\right)\)
\(A=25-xy\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(A=1.\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)