a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

a) ĐKXĐ: \(4x^3-9x\ne0\)

\(\Leftrightarrow x\left(4x^2-9\right)\ne0\)

\(\Leftrightarrow x\left(2x-3\right)\left(2x+3\right)\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\2x-3\ne0\\2x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\2x\ne3\\2x\ne-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne\frac{3}{2}\\x\ne-\frac{3}{2}\end{matrix}\right.\)

*Rút gọn

Ta có: \(M=\frac{\left(2x^3+3x^2\right)\left(2x+1\right)}{4x^3-9x}\)

\(=\frac{x^2\left(2x+3\right)\left(2x+1\right)}{x\left(4x^2-9\right)}=\frac{x^2\left(2x+3\right)\left(2x+1\right)}{x\left(2x+3\right)\left(2x-3\right)}=\frac{x\left(2x+1\right)}{2x-3}=\frac{2x^2+x}{2x-3}\)

Vậy: \(M=\frac{\left(2x^3+3x^2\right)\left(2x+1\right)}{4x^3-9x}=\frac{2x^2+x}{2x-3}\)

b) Khi M=0 thì

\(\frac{2x^2+x}{2x-3}=0\)

\(\Leftrightarrow2x^2+x=0\)

\(\Leftrightarrow x\left(2x+1\right)=0\)

\(\left[{}\begin{matrix}x=0\left(loại\right)\\2x+1=0\end{matrix}\right.\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)(tm)

Vậy: Khi M=0 thì \(x=\frac{-1}{2}\)

a)  M = ( 2x + 3)(2x - 3) - 2(x + 5)² - 2(x - 1)(x + 2) 

   = 4x² - 9 - 2(x² + 10x + 25) - 2(x² + x - 2)

   = 4x² - 9 - 2x² - 20x - 50 - 2x² - 2x + 4

   = -22x - 55 =  -11(2x + 5)

b) M = -11(2x + 5) = - 11(2.\(\frac{-7}{3}\)+ 5) = \(\frac{-11}{3}\)

b)  M = -11(2x + 5) = 0

\(\Rightarrow\)2x + 5 = 0

\(\Rightarrow\)x = \(\frac{-5}{2}\)

Ta có: M = (2x+3)(2x-3) - 2(x+5)² - 2(x-1)(x+2) \(=\left(2x\right)^2-3^2-2\left(x^2+10x+25\right)-\) \(2\left(x^2+x-2\right)\)

\(=4x^2-9-2x^2-20x-50-2x^2-2x+4\) =\(\left(4x^2-2x^2-2x^2\right)-\left(20x+2x\right)-\left(50+9-4\right)\) \(=-22x-55\)

b, Với x = \(-2\frac{1}{3}=\frac{-7}{3}\)

\(\Rightarrow M=-22.\frac{-7}{3}-55=\frac{154}{3}-55=\frac{-11}{3}\)

c, Để M = 0 => -22x - 55 = 0 \(\Rightarrow-22x=55\Rightarrow x=\frac{-55}{22}=\frac{-5}{2}\)

Vậy \(x=\frac{-5}{2}\)