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C =  \(\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\)\(\left(\frac{-\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)(  \(x\ge0\) , \(x\ne9;4\))

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

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

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

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

\(\frac{3}{\sqrt{x}-2}\)

#mã mã#

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

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

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

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

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

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

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

25 tháng 10 2016

\(C=\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{3x+3}{9-x}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{1}{2}\right)\) ĐK \(x\ge0;x\ne9\)

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

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

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

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

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

b: ta có \(C=\frac{-6}{\sqrt{x}+3}\)\(C=\frac{1}{2}\)

\(\frac{-6}{\sqrt{x}+3}=\frac{1}{2}\)

\(-12=\sqrt{x}+3\)

\(\sqrt{x}=-15\)(Loại)

=> x không có giá trị nào để C=\(\frac{1}{2}\)

28 tháng 6 2018

ĐKCĐ: \(x\ge0;x\ne9,x\ne4\)

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

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

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

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

b, \(A\inℤ\Leftrightarrow\frac{3}{\sqrt{x}-2}\inℤ\)

Nếu x không là số chính phương thì  \(\sqrt{x}\)là số vô tỉ thì \(\sqrt{x}-2\)là số vô tỉ\(\Rightarrow A=\frac{3}{\sqrt{x}-2}\)là số vô tỉ

Nếu x là số chính phương thì \(\sqrt{x}\)là số nguyên thì \(\sqrt{x}-2\inℤ\Rightarrow\sqrt{x}-2\inƯ\left(3\right)\Rightarrow\sqrt{x}-2\in\left\{\pm1;\pm3\right\}\Rightarrow\sqrt{x}\in\left\{1;3;5\right\}\)\(\Rightarrow x\in\left\{1;9;25\right\}\)

Mà theo ĐKXĐ có x khác 9 => \(x\in\left\{1,25\right\}\)

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

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

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

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