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27 tháng 7 2018

\(P=\left(\dfrac{x}{\sqrt{x}+3}-\dfrac{x+1}{\sqrt{x}-3}+\dfrac{6x+\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-1\right)\)

\(P=\dfrac{x\left(\sqrt{x}-3\right)-\left(x+1\right)\left(\sqrt{x}+3\right)+6x+\sqrt{x}}{x-9}:\dfrac{\left(\sqrt{x}-3\right)^2-x+9}{x-9}\)

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

\(P=-\dfrac{3}{x-9}:\dfrac{-6\sqrt{x}+18}{x-9}=-\dfrac{3}{x-9}.\dfrac{x-9}{-6\left(\sqrt{x}-3\right)}\)

\(P=\dfrac{1}{2\sqrt{x}-6}\)

b. \(x=12+6\sqrt{3}=9+2.3.\sqrt{3}+3=\left(3+\sqrt{3}\right)^2\Rightarrow\sqrt{x}=3+\sqrt{3}\)

\(P=\dfrac{1}{2.3+2\sqrt{3}-6}=\dfrac{1}{2\sqrt{3}}=\dfrac{\sqrt{3}}{6}\)

12 tháng 12 2022

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

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

b: Để A>-1 thì A+1>0

=>\(-3\sqrt{x}+2\sqrt{x}+4>0\)

=>-căn x>-4

=>0<x<16

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

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

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

 

1 tháng 8 2018

a, Rút gọn P

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1 tháng 8 2018

\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)

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

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

\(\Rightarrow\sqrt{x}+3\)

\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)

\(\Rightarrow\sqrt{y}-1\)

\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)

\(\Rightarrow\sqrt{xy}\)

1 tháng 8 2018

\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

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

\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)

\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)

\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)

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

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

17 tháng 7 2018

\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)

\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)

\(\Leftrightarrow\sqrt{x}-2< 0\)

\(\Leftrightarrow x< 4\)

Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)

KL............

\(2.\) Tương tự bài 1.

\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)

\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)

Bài 3: 

a: \(A=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{x-25}\)

\(=\dfrac{x-10\sqrt{x}+25}{x-25}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)

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

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

18 tháng 10 2018

Ukm ko để ý

14 tháng 10 2018

ĐỀ THI VÀO 10 ĐÓ CẢM ƠN MN TRƯỚC NHA:))