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30 tháng 4 2021

Ta có: \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt{xy}}\)

\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}\right)+\sqrt{y}\left(x-\sqrt{xy}\right)}{x^2-xy}\)

\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}+x-\sqrt{xy}\right)}{x\left(x-y\right)}=\dfrac{2x\sqrt{y}}{x\left(x-y\right)}\)

\(=\dfrac{2\sqrt{y}}{x-y}=\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(\Rightarrow A=\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}.\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\dfrac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\dfrac{\sqrt{x}+\sqrt{y}-1+1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)

AH
Akai Haruma
Giáo viên
3 tháng 8 2021

Bạn cần làm gì với biểu thức này?
 

\(F=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y}{xy}\cdot\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\dfrac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)

\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\dfrac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right]\)

\(=\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}\cdot xy=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

15 tháng 7 2023

\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{y-x}+\dfrac{1}{x+2\sqrt{x}\sqrt{y}+y}\right)-2x\) (với \(x\ne y,x,y\ge0\))

\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{1}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}+\dfrac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)

\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}+\dfrac{\sqrt{y}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}-\sqrt{x}\right)}\right)-2x\)

\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{\sqrt{y}+\sqrt{x}+\sqrt{y}-\sqrt{x}}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\right)-2x\)

\(P=\dfrac{4\sqrt{xy}}{x-y}:\left(\dfrac{2\sqrt{y}}{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)-2x\)

\(P=\dfrac{4\sqrt{xy}}{x-y}\cdot\dfrac{\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{2\sqrt{y}}-2x\)

\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\cdot2\sqrt{y}}-2x\)

\(P=\dfrac{4\sqrt{xy}\cdot\left(y-x\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\cdot2\sqrt{y}}-2x\)

\(P=\dfrac{2\sqrt{x}\left(y-x\right)}{\sqrt{x}-\sqrt{y}}-2x\)

\(P=\dfrac{2\sqrt{x}\left(y-x\right)-2x\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)

\(P=\dfrac{2y\sqrt{x}-2x\sqrt{x}-2x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

\(P=\dfrac{2y\sqrt{x}-4x\sqrt{x}+2x\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)

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: \(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)

2 tháng 10 2017

1.

\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)

hết tối giải rồi

24 tháng 7 2018

\(a.R=\left(\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}+\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)}{1-\sqrt{xy}}+1\right):\left(1-\dfrac{\sqrt{x}+1}{\sqrt{xy}+1}-\dfrac{\sqrt{x}\left(\sqrt{y}+1\right)}{\sqrt{xy}-1}\right)\)

\(R=\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)+xy-1}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]:\left[\dfrac{xy-1-\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)-\sqrt{x}\left(\sqrt{y}+1\right)\left(\sqrt{xy}+1\right)}{\left(\sqrt{xy}+1\right)\left(\sqrt{xy}-1\right)}\right]\)

\(R=\dfrac{x\sqrt{y}-\sqrt{x}+\sqrt{xy}-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}+xy-1}{xy-1}:\dfrac{xy-1-x\sqrt{y}+\sqrt{x}+\sqrt{xy}+1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}}{xy-1}\)

\(R=\dfrac{-2\sqrt{x}-2}{xy-1}:\dfrac{-2x\sqrt{y}-2\sqrt{xy}}{xy-1}\)

\(R=\dfrac{-2\left(\sqrt{x}+1\right)}{xy-1}.\dfrac{xy-1}{-2\left(x\sqrt{y}+\sqrt{xy}\right)}\)

\(R=\dfrac{\sqrt{x}+1}{x\sqrt{y}+\sqrt{xy}}\)

\(b.C=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{7\sqrt{x}+4}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)

\(C=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\dfrac{7\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(C=\dfrac{2x-6\sqrt{x}+7\sqrt{x}+4-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(C=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(C=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)

\(c.M=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+x}=\dfrac{\sqrt{x}+1+x}{x+\sqrt{x}}.\dfrac{\sqrt{x}+x}{\sqrt{x}}=\dfrac{\sqrt{x}+1+x}{\sqrt{x}}\)