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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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-x}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(A=\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)\(\div\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(A=\left(\frac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{2x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}}\)
\(A=\frac{2x+1}{4\sqrt{x}}\)
c, \(A=\frac{2x+1}{4\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\)
ap dụng cô si ta có \(\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\ge2\sqrt{\frac{\sqrt{x}}{2}\cdot\frac{1}{4\sqrt{x}}}=\frac{\sqrt{2}}{2}\)
dấu = xảy ra khi \(\frac{\sqrt{x}}{2}=\frac{1}{4\sqrt{x}}\Leftrightarrow x=\frac{1}{2}\) (tm)
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c) \(C=\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}=\)
\(C=\frac{x\sqrt{x}+2x+x+2\sqrt{x}-x\sqrt{x}+1}{\left(\left(\sqrt{x}\right)^3-1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{x-1}\times\frac{1}{x-1}=\frac{3x+2\sqrt{x}+1}{\left(x-1\right)^2}.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x}-1}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}-2}\right)\)
\(A=\left(\frac{\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\sqrt{x}}\right):\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(A=\left(\frac{\sqrt{x}-1}{x-\sqrt{x}}-\frac{\sqrt{x}}{x-\sqrt{x}}\right):\left(\frac{x-4}{x-2\sqrt{x}-\sqrt{x}+2}-\frac{x-1}{x-\sqrt{x}-2\sqrt{x}+2}\right)\)
\(A=\left(\frac{\sqrt{x}-1-\sqrt{x}}{x-\sqrt{x}}\right):\left(\frac{x-4-x+1}{x-3\sqrt{x}+2}\right)\)
\(A=\left(\frac{-1}{x-\sqrt{x}}\right):\left(\frac{-3}{2-2\sqrt{x}}\right)\)
\(A=\frac{-1\left(2-2\sqrt{x}\right)}{-3\left(x-\sqrt{x}\right)}=\frac{2\sqrt{x}-2}{3\sqrt{x}-3x}\)
Xog trục căn thức ở mẫu....
\(A=\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x}-1}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}-2}\right)\)(ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne4\end{matrix}\right.\))
\(=\frac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{x-2-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{-1}\)
\(=\frac{\sqrt{x}-2}{\sqrt{x}}\)
kết quả là :
A=P+x+19/9
ĐK : \(x\ge0\)
Ta có :
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right).\left(\frac{1}{\sqrt{x}+1}+\frac{1}{x-1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)\(.\left(\frac{1}{\sqrt{x}+1}+\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(x-1\right)\sqrt{x}}{\sqrt{x}\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{1}{\sqrt{x}+1}\)
Vậy ta có
\(A=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+19}{9}\)
\(=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+1}{9}+\frac{18}{9}\)
\(=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+1}{9}+2\)
Áp dụng BĐT Cauchy ta có
\(\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+1}{9}\ge2\sqrt{\frac{1}{\sqrt{x}+1}.\frac{\sqrt{x}+1}{9}}=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+1}{9}+2\ge\frac{8}{3}\)
\(\Leftrightarrow A\ge\frac{8}{3}\)
Dấu "=" xảy ra khi
\(\frac{1}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{9}\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=9\)
\(\Leftrightarrow\sqrt{x}+1=3\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)
Vậy GTNN của A là \(\frac{8}{3}\) đạt được khi x = 4