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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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
a. ĐKXĐ \(x\ge0\)và \(x\ne9\)
Ta có \(K=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. Để \(K< -1\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\Rightarrow4\sqrt{x}-6< 0\)vì \(\sqrt{x}+3\ge3\)
\(\Rightarrow0\le x< \frac{9}{4}\left(tm\right)\)
Vậy với \(0\le x< \frac{9}{4}\)thì K<-1
c. \(K=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta có \(\sqrt{x}+3\ge3\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\Rightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\)
\(\Rightarrow K\ge-3\)
Vậy \(MinK=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
\(ĐKXĐ:\hept{\begin{cases}x-4\ne0\\3-\sqrt{x}\ne0\\x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne4\\\sqrt{x}\ne3\\x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne4\\x\ne9\\x\ge0\end{cases}}\)
Rút gọn
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-1\right):\left(\frac{4-x}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-\frac{x-4}{x-4}\right):\left(\frac{4-x}{x+2\sqrt{x}-3\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{x-2\sqrt{x}-x+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-\left(x+4\sqrt{x}+4\right)-\left(x-6\sqrt{x}+9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-x^2-4\sqrt{x}-4-x^2+6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{-2x^2-x-2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\frac{\left(-2\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(-2x^2-x-2\sqrt{x}-9\right)}\)
\(D=\frac{\left(-2\right)\left(\sqrt{x}-3\right)\left(x^2-4\right)}{\left(x-4\right)\left(-2x^2-x-2\sqrt{x}-9\right)}\)
Sai thui nhé !!!!
Bài 1 :
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
b) Để \(A< -1\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)
\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)
\(\Leftrightarrow2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)
\(\Leftrightarrow x< \frac{1}{4}\)
Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)