\(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3...">
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\(=\dfrac{3\sqrt{x}-x+2x}{9-x}:\dfrac{\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

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

\(=\dfrac{x}{\sqrt{x}-5}\)

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

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

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

 

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

\(=\dfrac{3x-9\sqrt{x}+3\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{3x-6\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)

16 tháng 8 2018

Mình làm mấy bài rút gọn thôi nhé :v (mấy cái kia mình làm sợ không đúng)

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

\(=\dfrac{x+\sqrt{x}+1-\left(x+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}+1-x-2-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+1-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+0-x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

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

16 tháng 8 2018

Bài 3:

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

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

Bài 1:

a: \(A=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)

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

\(=\dfrac{3x+3\sqrt{x}}{9x-1}\cdot\dfrac{3\sqrt{x}+1}{3}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)

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

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

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

25 tháng 4 2017

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

b)\(x>36\)

27 tháng 5 2017

Ôn tập Căn bậc hai. Căn bậc ba

Ôn tập Căn bậc hai. Căn bậc ba

Bài 2:

a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)

\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)

\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)

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

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

24 tháng 7 2018

Ta có :

a , \(M=2\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\left[\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)

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

\(M=\left(\dfrac{2x-6\sqrt{x}-2x-18}{x-9}\right).\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\right]\)

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

\(M=\dfrac{-6\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)

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

b , mik ko chắc chắn nên mik chưa làm nhé !

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

\(=\sqrt{x}+2-\sqrt{x}-2=0\)

1 tháng 8 2018

a, Rút gọn P

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

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

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

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

\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{\left(\sqrt{x}+3\right).\left(3-\sqrt{x}\right).\left(x+\sqrt{3}\right).\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right).\left(2-\sqrt{x}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)

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

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

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

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

\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{\left(\sqrt[]{x}-2\right)^2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)

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

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

\(\Leftrightarrow3.\dfrac{1}{\sqrt{x}-2}\)

\(\Leftrightarrow\)\(\dfrac{3}{\sqrt{x}-2}\)