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27 tháng 8 2019

a, ĐKXĐ : \(x> 0 ; x \neq 1 \)

P = \(\dfrac{3x+3\sqrt{x} - 3}{\sqrt{x^2} +2\sqrt{x} - \sqrt{x} - 2}\) \(- \dfrac{\sqrt{x}+1}{\sqrt{x}+2} + \dfrac{\sqrt{x}-2}{\sqrt{x}} . \dfrac{1-( 1 -\sqrt{x})}{1-\sqrt{x}}\)

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

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

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

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

= \(\dfrac{3x+3\sqrt{x} - 3 - x+1-x+4}{(\sqrt{x}+2)(\sqrt{x}-1)} \)

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

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

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

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

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

c, Để P = \(\sqrt{x}\) \(\Leftrightarrow\) \(\dfrac{\sqrt{x}+1}{\sqrt{x} - 1} \) = \(\sqrt{x} \)

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

\(\Leftrightarrow\) \(\sqrt{x}+1 = \sqrt{x^2} - \sqrt{x}\)

\(\Leftrightarrow\) \( \sqrt{x^2} -\sqrt{x} - \sqrt{x} - 1 = 0\)

\(\Leftrightarrow\) \(\sqrt{x^2} - 2\sqrt{x} +1-1-1=0\)

\(\Leftrightarrow\) \((\sqrt{x}-1)^2 - (\sqrt{2})^2 \) = 0

\(\Leftrightarrow\) \((\sqrt{x} - 1 - \sqrt{2})(\sqrt{x} - 1+\sqrt{2})\)

\(\Leftrightarrow\) \(\begin{cases} \sqrt{x} - 1 - \sqrt{2}=0 \\ \sqrt{x} - 1 +\sqrt{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} \sqrt{x} = 1 +\sqrt{2} \\ \sqrt{x} = 1 - \sqrt{2} \end{cases} \) \(\Leftrightarrow\)\(\begin{cases} x = 1+\sqrt{2} = 3+2\sqrt{2} \\ \sqrt{x} = 1-\sqrt{2} < 0 ( LOẠI ) \end{cases} \)

P/s : mk không biết làm phần b

30 tháng 7 2019

\(ĐKXĐ:\)

\(\hept{\begin{cases}x-9\ne0\\\sqrt{x}-2\ne0\\\sqrt{x}+3\ne0;x\ge0\end{cases}}\)

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

Vậy...................................................

30 tháng 7 2019

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

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

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

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

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

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

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

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

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

9 tháng 2 2018

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

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

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

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

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

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

\(M=3\)

9 tháng 2 2018

b) \(\sqrt{x}=M\)

\(\Leftrightarrow x=M^2\)

thay vào ta có: 

\(x=3^2\)

\(x=9\)

c) \(M=3\in N\)

\(\Rightarrow x=3\)

d) \(M>1\Leftrightarrow x>1\)