Tìm x để
\(\frac{1}{\sqrt{x}}.\frac{1+x}{3\sqrt{x}}=1\)
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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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
9 tháng 2 2018
\(M=\frac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{\sqrt{x}}.\left(\frac{1}{1-\sqrt{x}}-1\right)\)
\(M=\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}-1}\)
\(M=\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{x-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\) \(+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(M=\frac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(M=\frac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(M=\frac{3\left(x+\sqrt{x}-2\right)}{x+\sqrt{x}-2}\)
\(M=3\)
12 tháng 8 2020
Ta có: \(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\frac{3-\sqrt{x}}{\sqrt{x}+1}\right)\) ( ĐKXĐ: \(x>0,\)\(x\ne0,\)\(x\ne1\))
\(\Leftrightarrow A=\left(\frac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow A=\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow A=\left(\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\right).\left(\frac{\sqrt{x}+1}{2.\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow A=\left(\frac{2\sqrt{x}}{\sqrt{x}}\right).\left(\frac{\sqrt{x}+1}{2.\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
Để \(A\ge\frac{3}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\ge\frac{3}{2}\)
Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-1}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{3}{2}\ge0\)
\(\Leftrightarrow\frac{2\sqrt{x}+2-3\sqrt{x}+3}{2.\left(\sqrt{x}-1\right)}\ge0\)
\(\Leftrightarrow\frac{5-\sqrt{x}}{2.\left(\sqrt{x}-1\right)}\ge0\)
+ TH1: \(\hept{\begin{cases}5-\sqrt{x}\ge0\\2\sqrt{x}-2\ge0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{x}\le5\\\sqrt{x}\ge1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\le25\\x\ge1\end{cases}}\)\(\Rightarrow\)\(1\le x\le25\)\(\left(TM\right)\)
+ TH2: \(\hept{\begin{cases}5-\sqrt{x}\le0\\2\sqrt{x}-2\le0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{x}\ge5\\\sqrt{x}\le1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\ge25\\x\le1\end{cases}}\)\(\left(L\right)\)
\(\Rightarrow\)\(1\le x\le25.\)Kết hợp ĐKXĐ: \(x\ne1\)
\(\Rightarrow\)\(1< x\le25\)
Vậy để \(A\ge\frac{3}{2}\)\(\Leftrightarrow\)\(1< x\le25\)