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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: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
1) Bạn đánh nhầm \(\sqrt{x}+3\rightarrow\sqrt{x+3}\); \(\sqrt{x}-3\rightarrow\sqrt{x-3}\)
Sửa : \(ĐKXĐ:x\ne\pm\sqrt{3}\)
a) \(M=\frac{x-\sqrt{x}}{x-9}+\frac{1}{\sqrt{x}+3}-\frac{1}{\sqrt{x}-3}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow M=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow M=\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
b) Để \(M=\frac{3}{4}\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}+3}=\frac{3}{4}\)
\(\Leftrightarrow4\sqrt{x}+8=3\sqrt{x}+9\)
\(\Leftrightarrow\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=1\)
\(\Leftrightarrow x=1\)(tm)
Vậy để \(A=\frac{3}{4}\Leftrightarrow x=1\)
c) Khi x = 4
\(\Leftrightarrow M=\frac{\sqrt{4}+2}{\sqrt{4}+3}\)
\(\Leftrightarrow M=\frac{2+2}{2+3}\)
\(\Leftrightarrow M=\frac{4}{5}\)
Vậy khi \(x=4\Leftrightarrow M=\frac{4}{5}\)
a) \(B=\left(\sqrt{x}-\frac{9}{\sqrt{x}}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}}-\frac{9\sqrt{x}+9}{x+3\sqrt{x}}\right)\)
\(B=\frac{x-9}{\sqrt{x}}:\left(\frac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}\left(\sqrt{x}+3\right)}-\frac{9\sqrt{x}+9}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(B=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}}\cdot\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x+6\sqrt{x}+9-9\sqrt{x}-9}\)
\(B=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)^2}{x-3\sqrt{x}}\)
\(B=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)^2}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(B=\frac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}}\)
b) \(2B=\sqrt{x}+31\)
\(\Leftrightarrow\frac{2\left(\sqrt{x}+3\right)^2}{\sqrt{x}}=\sqrt{x}+31\)
\(\Leftrightarrow2\left(x+6\sqrt{x}+9\right)=\sqrt{x}\left(\sqrt{x}+31\right)\)
\(\Leftrightarrow2x+12\sqrt{x}+18=x+31\sqrt{x}\)
\(\Leftrightarrow x-19\sqrt{x}+18=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-18=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=324\end{matrix}\right.\)( thỏa )
Vậy....
c) \(M=B-\frac{5}{\sqrt{x}}\)
\(M=\frac{\left(\sqrt{x}+3\right)^2-5}{\sqrt{x}}\)
\(M=\frac{x+6\sqrt{x}+9-5}{\sqrt{x}}\)
\(M=\frac{x+6\sqrt{x}+4}{\sqrt{x}}\)
\(M=\sqrt{x}+6+\frac{4}{\sqrt{x}}\)
Đặt \(\frac{1}{\sqrt{x}}=a\)
Áp dụng bất đẳng thức Cô-si :
\(M=\frac{1}{a}+6+4a\ge2\sqrt{\frac{4a}{a}}+6=10\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{1}{a}=4a\Leftrightarrow a=\frac{1}{2}\Leftrightarrow\frac{1}{\sqrt{x}}=\frac{1}{2}\Leftrightarrow x=4\)( thỏa )
Vậy....