\(P=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}+\frac{\sqrt{x}}{2+\sqrt{x}}-\frac{4x+2\sqrt{x}-4}{x-4}\right):\left(\frac{2}{2-\sqrt{x}}-\frac{\sqrt{x}+3}{2\sqrt{x}-x}\right)\)
Tìm x để P>0 Tìm x để P=-1
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5 tháng 2 2022
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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
3 tháng 2 2022
\(=\left(\dfrac{-\left(\sqrt{x}+2\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{4x}{x-4}\right)\cdot\dfrac{2\sqrt{x}-x}{\sqrt{x}-3}\)
\(=\dfrac{-x-4\sqrt{x}-4+x-4\sqrt{x}+4-4x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-3}\)
\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{-\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{4x}{\sqrt{x}-3}\)
Để P>0 thì \(\sqrt{x}-3>0\)
hay x>9
Để P<0 thì \(\sqrt{x}-3< 0\)
hay 0<x<9
13 tháng 8 2020
a) Ta có: \(P=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}+\frac{\sqrt{x}}{2+\sqrt{x}}-\frac{4x+2\sqrt{x}-4}{x-4}\right):\left(\frac{2}{2-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4x+2\sqrt{x}-4}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{-2}{\sqrt{x}-2}-\frac{\sqrt{x}-3}{\sqrt{x}+3}\right)\)
\(=\frac{4+4\sqrt{x}+x+2\sqrt{x}-x+4x+2\sqrt{x}-4}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\left(\frac{-2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{4x+8\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{-2\sqrt{x}-6-\left(x-5\sqrt{x}+6\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(2-\sqrt{x}\right)\left(\sqrt{x}+2\right)}:\frac{-2\sqrt{x}-6-x+5\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}}{\sqrt{x}-2}:\frac{-x+3\sqrt{x}-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{-4\sqrt{x}}{\sqrt{x}-2}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-x+3\sqrt{x}-12}\)
\(=\frac{-4\sqrt{x}\left(\sqrt{x}+3\right)}{-x+3\sqrt{x}-12}\)
\(=\frac{4x+12\sqrt{x}}{x-3\sqrt{x}+12}\)
b)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
Để P=-1 thì \(\frac{4x+12\sqrt{x}}{x-3\sqrt{x}+12}=-1\)
\(\Leftrightarrow4x+12\sqrt{x}=-1\left(x-3\sqrt{x}+12\right)\)
\(\Leftrightarrow4x+12\sqrt{x}=-x+3\sqrt{x}-12\)
\(\Leftrightarrow4x+12\sqrt{x}+x-3\sqrt{x}+12=0\)
\(\Leftrightarrow5x+9\sqrt{x}+12=0\)(1)
Ta có: \(\forall x\) thỏa mãn ĐKXĐ ta luôn có: \(\left\{{}\begin{matrix}5x\ge0\\9\sqrt{x}\ge0\end{matrix}\right.\Leftrightarrow5x+9\sqrt{x}\ge0\Leftrightarrow5x+9\sqrt{x}+12>0\)(2)
Từ (1) và (2) suy ra không có giá trị nào của x để P=-1