a ) \(5x^2\left(2x-3\right)+\left(2x^2+3x+3\right)\left(3-2x\right)=6x^3-9x^2\)
\(\Rightarrow5x^2\left(2x-3\right)-\left(2x^2+3x+3\right)\left(2x-3\right)=3x^2\left(2x-3\right)\)
\(\Rightarrow5x^2\left(2x-3\right)-\left(2x^2+3x+3\right)\left(2x-3\right)-3x^2\cdot\left(2x-3\right)=0\)
\(\Rightarrow\left(5x^2-2x^2-3x-3-2x+3\right)\left(2x-3\right)=0\)
\(\Rightarrow\left(3x^2-5x\right)\left(2x-3\right)=0\)
\(\Rightarrow x\left(3x-5\right)\left(2x-3\right)=0\)
\(\Rightarrow\) +) x = 0
+) 3x - 5 = 0\(\Rightarrow x=\dfrac{5}{3}\)
+ )\(2x-3=0\Rightarrow x=\dfrac{3}{2}\)
vậy x \(=0;x=\dfrac{3}{2};x=\dfrac{5}{3}\)
b) \(8x^3+12x^2+6x+7-3\left(2x+1\right)^2=6\)
\(\Rightarrow\left(2x\right)^3+3.2x.1+3.2x.1^2+1^2+6-3\left(2x+1\right)^2-6=0\)
\(\Rightarrow\left(2x+1\right)^3-3\left(2x+1\right)^2=0\)
\(\Rightarrow\left(2x+1\right)^2\left(2x+1-3\right)=0\)
\(\Rightarrow\left(2x+1\right)^2\left(2x-2\right)=0\Rightarrow\left(2x+1\right)^2\left(x-1\right)2=0\)
\(\Rightarrow\) +)\(\left(2x+1\right)^2=0\Rightarrow2x+1=0\Rightarrow x=\dfrac{-1}{2}\)
+) x - 1 = 0 \(\Rightarrow x=1\)
Vậy x = \(\dfrac{-1}{2}\) hoặc x = 1