ĐK: \(x>9;x\ne1\)
a) \(B=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-1}+1\right)\)
\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{2\sqrt{x}-2+\sqrt{x}-1}{\sqrt{x}-1}\)
\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}:\frac{3\sqrt{x}-3}{\sqrt{x}-1}\)
\(B=\frac{-3\sqrt{x}-3}{x-9}\cdot\frac{\sqrt{x}-1}{3\left(\sqrt{x}-1\right)}\)
\(B=\frac{-3\left(\sqrt{x}+1\right)}{3\left(x-9\right)}\)
\(B=\frac{\sqrt{x}+1}{9-x}\)
b) Để B nguyên mà \(9-x\) nguyên nên:
\(\sqrt{x}+1⋮9-x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)⋮9-x\)
\(\Leftrightarrow x-1⋮9-x\)
\(\Leftrightarrow-\left(9-x\right)+8⋮9-x\)
\(\Leftrightarrow8⋮9-x\)
\(\Leftrightarrow9-x\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Leftrightarrow x\in\left\{8;10;7;11;5;13;1;17\right\}\)
Mà theo ĐKXĐ nên \(x\in\left\{10;11;13;17\right\}\)( nhớ thử lại )
Vậy...
c) \(B=\frac{\sqrt{x}+1}{9-x}\)
\(\Leftrightarrow B\left(9-x\right)=\sqrt{x}+1\)
\(\Leftrightarrow9B-Bx=\sqrt{x}+1\)
\(\Leftrightarrow xB+\sqrt{x}-\left(9B-1\right)=0\)
\(\Delta=1^2+4\cdot B\cdot\left(9B-1\right)\ge0\)
\(\Leftrightarrow1+36B^2-4B\ge0\)
\(\Leftrightarrow\left(6B\right)^2-2\cdot6B\cdot\frac{1}{3}+\frac{1}{9}+\frac{8}{9}\ge0\)
\(\Leftrightarrow\left(6B-\frac{1}{3}\right)^2+\frac{8}{9}\ge0\)( luôn đúng )
Vậy B không có giá trị nhỏ nhất.
d) \(x=25-4\sqrt{6}\)
\(\Leftrightarrow x=\left(2\sqrt{6}\right)^2-2\cdot2\sqrt{6}\cdot1+1=\left(2\sqrt{6}-1\right)^2\)
\(\Leftrightarrow\sqrt{x}=2\sqrt{6}-1\)
Khi đó : \(B=\frac{2\sqrt{6}-1+1}{9-25+4\sqrt{6}}=\frac{2\sqrt{6}}{-16+4\sqrt{6}}=\frac{\sqrt{6}}{-8+2\sqrt{6}}\)
Vậy...
