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21 tháng 8 2023

C= \(\dfrac{√x-√y}{xy√xy}\) : 

\(\dfrac{(1}{x})\) + \(\dfrac{(1}{y)}\) . \(\dfrac{1}{x+y+2√xy}\)

+ \(\dfrac{2}{(√x+√y)³}\) . \(\dfrac{(1}{√x)}\) + \(\dfrac{(1}{√y)}\) 

Tính C với x = 2 - √3 ; y = 2+√3

a: \(\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+y+2\sqrt{xy}}\)

\(=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\frac{x+y}{xy}\cdot\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)

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

\(\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\)

\(=\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\cdot\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

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

Ta có: \(C=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+y+2\sqrt{xy}}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\)

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

b: \(\left(\sqrt{x}+\sqrt{y}\right)^2=\left(\sqrt{2-\sqrt3}+\sqrt{2+\sqrt3}\right)^2\)

\(=2-\sqrt3+2+\sqrt3+2\cdot\sqrt{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}=4+2=6\)

\(\sqrt{xy}=\sqrt{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}=\sqrt{4-3}=1\)

\(x+y=2+\sqrt3+2-\sqrt3=4\)

\(\sqrt{x}-\sqrt{y}=\sqrt{2-\sqrt3}-\sqrt{2+\sqrt3}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{4-2\sqrt3}-\sqrt{4+2\sqrt3}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt3-1\right)^2}-\sqrt{\left(\sqrt3+1\right)^2}\right)=\frac{1}{\sqrt2}\left(\sqrt3-1-\sqrt3-1\right)=-\frac{2}{\sqrt2}=-\sqrt2\)

Ta có: \(C=\frac{2\left(x+y\right)+\sqrt{x}-\sqrt{y}}{\sqrt{xy}\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(=\frac{2\cdot4-\sqrt2}{1\cdot4\cdot6}=\frac{8-\sqrt2}{24}\)

AH
Akai Haruma
Giáo viên
3 tháng 8 2021

Bạn cần làm gì với biểu thức này?
 

4 tháng 9 2022

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

30 tháng 4 2021

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

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

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

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

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

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

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