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25 tháng 7 2022

đk x>=0 ; x khác 4 ; 9 

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

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

a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)

\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)

b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)

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

\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)

g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)

\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)

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

 

28 tháng 7 2018

Bạn ơi đề bài là jlolang

a: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\dfrac{-3+2\sqrt{x}+2}{x\sqrt{x}+1}\)

\(=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\dfrac{2\sqrt{x}-1}{x\sqrt{x}+1}\)

\(=\dfrac{2x^2+2\sqrt{x}-9x\sqrt{x}-9+2x\sqrt{x}-10x+12\sqrt{x}-x+5\sqrt{x}-6}{\left(x\sqrt{x}+1\right)\left(x-5\sqrt{x}+6\right)}\)

\(=\dfrac{2x^2+19\sqrt{x}-7x\sqrt{x}-11x-15}{\left(x\sqrt{x}+1\right)\left(x-5\sqrt{x}+6\right)}\)

b: \(=\dfrac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{x\sqrt{x}+1}\)

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

1 tháng 1 2019

a)\(=4\sqrt{6}-3\sqrt{6}+1-\sqrt{6}\)

\(=1\)

b)ĐK: \(x>0,x\ne9\)

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

\(=\dfrac{3\sqrt{x}-9}{\sqrt{x}+3}.\dfrac{\sqrt{x}}{2\sqrt{x}+4}\)

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

1 tháng 8 2018

\(\dfrac{2\sqrt{X}-9}{x-5\sqrt{X}+6}-\dfrac{\sqrt{X}+3}{\sqrt{X}-2}-\dfrac{2\sqrt{X}+1}{3-\sqrt{X}}\) \(\left(X\ne2;X\ne3,X\ge0\right)\)

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

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

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

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

\(C=\dfrac{\sqrt{X}+1}{\sqrt{X}-3}< 1\)

\(\Rightarrow\dfrac{\sqrt{X}+1-\sqrt{X}+3}{\sqrt{X}-3}< 0\)

\(\Rightarrow\dfrac{4}{\sqrt{X}+3}< 0\) ( VÔ LÍ)

Không có X thỏa mãn

1 tháng 8 2018

sai dấu tỷ ới , hình như là \(\dfrac{4}{\sqrt{X}-3}< 0\)

30 tháng 7 2018

4 , Ta có :

\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x-9}{x-9}\)

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

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x+9}{x-9}\)

\(=\dfrac{3\sqrt{x}+9}{x-9}\)

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

\(=\dfrac{3}{\sqrt{x}-3}\)

30 tháng 7 2018

2 , Ta có :

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

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

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

a: \(=-4+2\sqrt{5}-\sqrt{5}+2+\sqrt{5}=2\sqrt{5}-2\)

b: \(B=\dfrac{2\sqrt{x}+4+6\sqrt{x}-3-2\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}}{6\sqrt{x}+4}\)

\(=\dfrac{\left(6\sqrt{x}+1\right)\cdot\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\left(6\sqrt{x}+4\right)}\)

24 tháng 2 2019

\(Q= \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{2\sqrt x + 1}}{{3 - \sqrt x }}\\\)
\(Q = \dfrac{{2\sqrt x - 9}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} + \dfrac{{2\sqrt x + 1}}{{\sqrt x - 3}}\\ Q = \dfrac{{2\sqrt x - 9 - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\ Q = \dfrac{{ - \sqrt x + x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\)

2 tháng 9 2018

\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{2\sqrt{x}-9+2x-4\sqrt{x}+\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

28 tháng 9 2017

mình sửa chút chỗ cuối ko có 1đâu nhéhaha

28 tháng 9 2017

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