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....
