Rút gọn:
a. \(\dfrac{2}{x^2-y^2}\sqrt{\dfrac{3\left(x+y\right)^2}{2}}\) với \(x\ge0,y\ge0\) và \(x\ne y;\)
b. \(\dfrac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\) với a > 0,5.
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a: \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+1\)
=2
c: \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
d: \(\dfrac{y-2\sqrt{y}+1}{\sqrt{y}-1}=\sqrt{y}-1\)
1: \(A=\dfrac{x-2\sqrt{xy}+y}{x-y}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
2: Thay \(x=3+2\sqrt{2}\) và \(y=3-2\sqrt{2}\) vào A, ta được:
\(A=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{\sqrt{2}+1+\sqrt{2}-1}=\dfrac{2}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
a) ta có : \(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
b) ta có : \(\dfrac{x-\sqrt{3x}+3}{x\sqrt{x}+3\sqrt{3}}=\dfrac{x-\sqrt{3x}+3}{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3x}+3\right)}=\dfrac{1}{\sqrt{x}+\sqrt{3}}\)
c) ta có : \(\dfrac{2}{\sqrt{5}-\sqrt{3}}+\dfrac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\dfrac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}+\dfrac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\) \(=\sqrt{5}+\sqrt{3}+\sqrt{6}-\sqrt{3}=\sqrt{5}+\sqrt{6}\)
Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
a) Rút gọn được \(\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
c) \(H=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\Rightarrow H^2=\dfrac{xy}{\left(x-\sqrt{xy}+y\right)^2}\)
\(\Rightarrow H^2-H=\dfrac{xy}{\left(x-\sqrt{xy}+y\right)^2}-\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}=\dfrac{xy-\sqrt{xy}\left(x-\sqrt{xy}+y\right)}{\left(x-\sqrt{xy}+y\right)^2}\)
\(=\dfrac{2xy-x\sqrt{xy}-y\sqrt{xy}}{\left(x-\sqrt{xy}+y\right)^2}=\dfrac{-\sqrt{xy}\left(x-2\sqrt{xy}+y\right)}{\left(x-\sqrt{xy}+y\right)^2}=-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x-\sqrt{xy}+y\right)^2}\)
Do \(\left\{{}\begin{matrix}\sqrt{xy}\ge0\\\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\\\left(x-\sqrt{xy}+y\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow H^2-H=-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x-\sqrt{xy}+y\right)^2}\le0\Rightarrow H^2\le H\)
Mà \(H\ge0\left(cmt\right)\Rightarrow H\le\sqrt{H}\)
a) \(B=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x,y\ge0;x\ne y\right)\)
\(B=\left[\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{x-y}\right]:\dfrac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)
\(B=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}}{x+\sqrt{xy}+y}\)
b) Xét tử:
\(\sqrt{xy}\ge0\forall x,y\) (xác định) (1)
Xét mẫu:
\(x+\sqrt{xy}+y\)
\(=\left(\sqrt{x}\right)^2+2\cdot\dfrac{1}{2}\sqrt{y}\cdot\sqrt{x}+\left(\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)
\(=\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)
Mà: \(\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2\ge0\forall x,y\) (xác định), còn: \(\dfrac{3}{4}y\ge0\) vì theo đkxđ thì \(y\ge0\) (2)
Từ (1) và (2) ⇒ B luôn không âm với mọi x,y (\(B\ge0\)) (đpcm)
a) Vì
nên
. Do đó:
=![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?%5Cfrac%7B2%7D%7Bx-y%7D%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%3D%5Cfrac%7B1%7D%7Bx-y%7D%5Csqrt%7B%5Cfrac%7B4%5Ccdot%203%7D%7B2%7D%7D%3D%5Cfrac%7B%5Csqrt%7B6%7D%7D%7Bx-y%7D.)
b)![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?%5Cfrac%7B2%7D%7B2a-1%7D%5Csqrt%7B5a%5E%7B2%7D%281-4a+4a%5E%7B2%7D%29%7D%3D%5Cfrac%7B2%7D%7B2a-1%7D%5Csqrt%7B5a%5E%7B2%7D%282a-1%29%5E%7B2%7D%7D%3D%5Cfrac%7B2%5Cleft%20%7C%20a%20%5Cright%20%7C%5Ccdot%20%5Cleft%20%7C%202a-1%20%5Cright%20%7C%7D%7B2a-1%7D%5Csqrt%7B5%7D.)
Vì a>0,5 nên 2a-1>0. Do đó
.