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24 tháng 11 2021

\(\Leftrightarrow\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)\(-\)\(\frac{\sqrt{x}+3}{\sqrt{x}-2}\)\(+\)\(\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(\Leftrightarrow\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)\(-\)\(\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)\(+\)\(\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow\frac{2\sqrt{x}-9-x+3\sqrt{x}-3\sqrt{x}+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow\frac{-\sqrt{x}+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

19 tháng 9 2020

Mình tách thành hai phần nhìn cho dễ hiểu nhé !

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

+) \(\frac{x-3\sqrt{x}}{x-9}-1=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}-1=\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{-3}{\sqrt{x}+3}\)

+) \(\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\)

\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

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

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

=> \(\frac{-3}{\sqrt{x}+3}\div\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{-3}{\sqrt{x}+3}\times\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{4-x}\)

\(=\frac{3\left(\sqrt{x}-2\right)}{x-4}=\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3}{\sqrt{x}+2}\)

6 tháng 7 2017

\(\left(\frac{x+3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

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

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

\(=\frac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\frac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-3}{\sqrt{x}+3}:\frac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

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

\(=\frac{-3\cdot\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}\)

\(=\frac{-3}{-\left(\sqrt{x}-2\right)}=\frac{3}{\sqrt{x}-2}\)

Chúc bạn học giỏi 

Kết bạn với mình nha

\(P=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

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

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-x+9+2x-4\sqrt{x}-5}\)

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

23 tháng 5 2021

Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé

8 tháng 7 2019
Bạn xem lại có ghi sai đề ko vậy
8 tháng 7 2019

đề không sai đâu bạn

11 tháng 7 2018

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)

\(=1-a+a=1\)