a: \(2x^3+3x^2-3x-2=0\)
=>\(\left(2x^3-2\right)+\left(3x^2-3x\right)=0\)
=>\(2\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(2x^2+2x+2+3x\right)=0\)
=>\(\left(x-1\right)\left(2x^2+4x+x+2\right)=0\)
=>(x-1)(x+2)(2x+1)=0
=>\(\left[{}\begin{matrix}x=1\\x=-2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: \(x^3+1-\left(x+1\right)\left(x^2-5\right)=0\)
=>\(\left(x+1\right)\left(x^2-x+1\right)-\left(x+1\right)\left(x^2-5\right)=0\)
=>\(\left(x+1\right)\left(x^2-x+1-x^2+5\right)=0\)
=>(x+1)(6-x)=0
=>\(\left[{}\begin{matrix}x=-1\\x=6\end{matrix}\right.\)
c: \(x^3+2x^2-x-2=0\)
=>\(x^2\left(x+2\right)-\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2-1\right)=0\)
=>(x+2)(x-1)(x+1)=0
=>\(\left[{}\begin{matrix}x=-2\\x=-1\\x=1\end{matrix}\right.\)
d: \(x^2-x-2=0\)
=>\(x^2-2x+x-2=0\)
=>(x-2)(x+1)=0
=>\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
e: Sửa đề: \(x^2+2x-15=0\)
=>\(x^2+5x-3x-15=0\)
=>x(x+5)-3(x+5)=0
=>(x+5)(x-3)=0
=>\(\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
f: \(4x^2-12x+5=0\)
=>\(4x^2-10x-2x+5=0\)
=>2x(2x-5)-(2x-5)=0
=>(2x-5)(2x-1)=0
=>\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
