m: \(2x^3-x^2+3x+6=0\)
=>\(2x^3+2x^2-3x^2-3x+6x+6=0\)
=>\(\left(x+1\right)\left(2x^2-3x+6\right)=0\)
=>x+1=0
=>x=-1
o: \(\left(x^2-4x\right)^2-8\left(x^2-4x\right)+15=0\)
=>\(\left(x^2-4x\right)^2-3\left(x^2-4x\right)-5\left(x^2-4x\right)+15=0\)
=>\(\left(x^2-4x-3\right)\left(x^2-4x-5\right)=0\)
TH1: \(x^2-4x-3=0\)
=>\(x^2-4x+4-7=0\)
=>\(\left(x-2\right)^2=7\)
=>\(\left[\begin{array}{l}x-2=\sqrt7\\ x-2=-\sqrt7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt7+2\\ x=-\sqrt7+2\end{array}\right.\)
TH2: \(x^2-4x-5=0\)
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
b: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
=>\(x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=15\)
=>\(x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\)
=>24x+25=15
=>24x=-10
=>\(x=-\frac{10}{24}=-\frac{5}{12}\)
