cho \(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\frac{3-\sqrt{x}}{\sqrt{x}+1}\right).\)
Tìm x để A \(\ge\frac{3}{2}\)
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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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
22 tháng 6 2016
c) \(C=\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}=\)
\(C=\frac{x\sqrt{x}+2x+x+2\sqrt{x}-x\sqrt{x}+1}{\left(\left(\sqrt{x}\right)^3-1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{x-1}\times\frac{1}{x-1}=\frac{3x+2\sqrt{x}+1}{\left(x-1\right)^2}.\)
17 tháng 10 2018
\(a)\)\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt{x-3}}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}-3}{\sqrt{x}-3}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(R=\frac{3\sqrt{x}+3}{\sqrt{x}+3}.\frac{\sqrt{x}-3}{\sqrt{x+1}}\)
\(R=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
\(b)\) Ta có : \(R< -1\)
\(\Leftrightarrow\)\(\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}< -1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-3}{\sqrt{x}+3}< \frac{-1}{3}\)
\(\Leftrightarrow\)\(3\sqrt{x}-9< -\sqrt{x}-3\)
\(\Leftrightarrow\)\(4\sqrt{x}< 6\)
\(\Leftrightarrow\)\(\sqrt{x}< \frac{3}{2}\)
\(\Leftrightarrow\)\(x< \frac{9}{4}\)
Chúc bạn học tốt ~
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 \(A\ge\frac{3}{2}\)\(\Leftrightarrow\)\(1< x\le25\)