Vd7:
a: Ta có: \(M=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
b: Thay \(x=11-6\sqrt{2}\) vào M, ta được:
\(M=\dfrac{3+\sqrt{2}+1}{3+\sqrt{2}-3}=\dfrac{4+\sqrt{2}}{\sqrt{2}}=2\sqrt{2}+1\)
c: Để M=2 thì \(\sqrt{x}+1=2\sqrt{x}-6\)
\(\Leftrightarrow-\sqrt{x}=-7\)
hay x=49
Vd7:
d: Để M<1 thì M-1<0
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)
e: Để M nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow4⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-2;-1;1;2;4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{1;4;5;7\right\}\)
hay \(x\in\left\{1;16;25;49\right\}\)
