cho biểu thức
\(P=\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\)
a) rút gọn biểu thức P
b)tính giá trị của biểu thức P với x=\(\dfrac{2}{2+\sqrt{3}}\)
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12 tháng 10 2022
a: \(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{x-1}=\dfrac{-2\left(\sqrt{x}-1\right)}{x-1}=\dfrac{-2}{\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}:\left(\dfrac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
c: \(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
AT
20 tháng 6 2021
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) \(x=\left(1-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)
\(y=3-\sqrt{8}\Rightarrow\sqrt{y}=\sqrt{3-\sqrt{8}}=\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)
\(\Rightarrow M=\left(\sqrt{3}-1\right)-\left(\sqrt{2}-1\right)=\sqrt{3}-\sqrt{2}\)
30 tháng 4 2021
Ta có: \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt{xy}}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}\right)+\sqrt{y}\left(x-\sqrt{xy}\right)}{x^2-xy}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}+x-\sqrt{xy}\right)}{x\left(x-y\right)}=\dfrac{2x\sqrt{y}}{x\left(x-y\right)}\)
\(=\dfrac{2\sqrt{y}}{x-y}=\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}.\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1+1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)
5 tháng 7 2021
a) Ta có: \(P=\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+2xy+y}{1-xy}\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{1-xy+x+2xy+y}{1-xy}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}\cdot\dfrac{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}{x+xy+y+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
5 tháng 7 2021
Đk:\(xy\ne1;x\ge0;y\ge0\)
\(P=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{1-xy+x+y+2xy}{1-xy}\)
\(=\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{1+x+y+xy}{1-xy}\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{\left(1+x\right)\left(1+y\right)}{1-xy}\)\(=\dfrac{2\sqrt{x}\left(1+y\right)}{1-xy}.\dfrac{1-xy}{\left(1+x\right)\left(1+y\right)}=\dfrac{2\sqrt{x}}{1+x}\)
b) Áp dụng AM-GM có:
\(1+x\ge2\sqrt{x}\Leftrightarrow\)\(\dfrac{2\sqrt{x}}{1+x}\le1\)
Dấu "=" xảy ra khi x=1 (tm)
Vậy \(P_{max}=1\)
\(\dfrac{\left(\sqrt{X}+\sqrt{Y}\right)\left(1+\sqrt{XY}\right)+\left(\sqrt{X}-\sqrt{Y}\right)\left(1-\sqrt{XY}\right)}{1-XY}\cdot\dfrac{1-XY}{1-XY+\sqrt{X}+\sqrt{Y}+2\sqrt{XY}}=\dfrac{\sqrt{X}+X\sqrt{Y}+\sqrt{Y}+Y\sqrt{X}+\sqrt{X}-X\sqrt{Y}-\sqrt{Y}+Y\sqrt{X}}{1-XY}\cdot\dfrac{1-XY}{XY+X+Y+1}=\dfrac{2\sqrt{X}\left(1+Y\right)}{\left(1+Y\right)\left(X+1\right)}=\dfrac{2\sqrt{X}}{X+1}\)
b: Thay \(x=\dfrac{2}{2+\sqrt{3}}=2\left(2-\sqrt{3}\right)=4-2\sqrt{3}\) vào P, ta được:
\(P=\dfrac{2\left(\sqrt{3}-1\right)}{4-2\sqrt{3}+1}=\dfrac{2\sqrt{3}-2}{5-2\sqrt{3}}=\dfrac{6\sqrt{3}+2}{13}\)