Bài 1:
a)
\(M=\frac{2x}{x+3}+\frac{x+1}{x-3}+\frac{11x-3}{x^2-9}\\ \Leftrightarrow M=\frac{2x\cdot\left(x-3\right)}{x^2-9}+\frac{\left(x+1\right)\cdot\left(x+3\right)}{x^2-9}+\frac{11x-3}{x^2-9}\\ \Leftrightarrow M=\frac{2x^2-6x}{x^2-9}+\frac{x^2+4x+3}{x^2-9}+\frac{11x-3}{x^2-9}\\ \Leftrightarrow M=\frac{2x^2-6x+x^2+4x+3+11x-3}{x^2-9}\\ \Leftrightarrow M=\frac{3x^2+9x}{x^2-9}\\ \Leftrightarrow M=\frac{3x\cdot\left(x+3\right)}{\left(x+3\right)\cdot\left(x-3\right)}\\ \Rightarrow M=\frac{3x}{x-3}\)
b) Theo đề bài có:
\(M=\frac{3}{2}\\ \Leftrightarrow M=\frac{3}{x-3}=\frac{3}{2}\\ \Rightarrow x-3=\frac{3\cdot2}{3}=2\\ \Rightarrow x=2+3=5\\ \Rightarrow x=5\)
Vậy x = 5 thì \(M=\frac{3}{2}\)
c) Theo đề bài có:
\(\frac{1}{M}\le\frac{1}{6}\\ \Leftrightarrow\frac{1}{\frac{3x}{x-3}}=\frac{1}{3x^2-9x}\le\frac{1}{6}\\ \Rightarrow3x^2-9x\le\frac{6\cdot1}{1}\le6\\ \Leftrightarrow x^2-3x\le2\\ \Leftrightarrow\left(x-1.5\right)\cdot\left(x-15\right)\le4.25\\ \Rightarrow x-1.5\le\frac{\sqrt{17}}{2}\\ \Rightarrow x\le\frac{\sqrt{17}}{2}+1.5\le\frac{3+\sqrt{17}}{2}\)
Vậy \(x\le\frac{3+\sqrt{17}}{2}\) thì \(\frac{1}{M}\le\frac{1}{6}\)
