tìm x
a, \(\left(x+2\right)^3-\left(x+1\right)\left(x^2-1+1\right)-6\left(x-1\right)^2=23\)
b, \(\left(x+3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)+11=0\)
c, \(x\left(x-3\right)-x+3=0\)
d, \(\left(x-1\right)\left(x+2\right)-2x-4=0\)
e, \(x^3-3x^2-4x+12=0\)
f, \(9\left(x-1\right)-x^3+x^2=0\)
g, \(x^3-2x^2+x-2=0\)
h, \(\left(9-x^2\right)\left(2x+1\right)=0\)
a: Sửa đề: \(\left(x+2\right)^3-\left(x+1\right)\left(x^2-x+1\right)-6\left(x-1\right)^2=23\)
=>\(x^3+6x^2+12x+8-\left(x^3-1\right)-6\left(x^2-2x+1\right)=23\)
=>\(x^3+6x^2+12x+8-x^3+1-6x^2+12x-6=23\)
=>24x+3=23
=>24x=20
=>\(x=\dfrac{20}{24}=\dfrac{5}{6}\)
b: \(\left(x+3\right)\left(x^2+3x+9\right)-x\left(x-2\right)\left(x+2\right)+11=0\)
=>\(x^3+3x^2+9x+3x^2+9x+27-x\left(x^2-4\right)+11=0\)
=>\(x^3+6x^2+18x+27-x^3+4x+11=0\)
=>\(6x^2+22x+38=0\)
\(\Delta=22^2-4\cdot6\cdot38=-428< 0\)
=>Phương trình vô nghiệm
c: x(x-3)-x+3=0
=>x(x-3)-(x-3)=0
=>(x-3)(x-1)=0
=>\(\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
d: \(\left(x-1\right)\left(x+2\right)-2x-4=0\)
=>\(\left(x-1\right)\left(x+2\right)-2\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
e: \(x^3-3x^2-4x+12=0\)
=>\(x^2\left(x-3\right)-4\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x^2-4\right)=0\)
=>(x-3)(x-2)(x+2)=0
=>\(\left[{}\begin{matrix}x-3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=-2\end{matrix}\right.\)
f: \(9\left(x-1\right)-x^3+x^2=0\)
=>\(9\left(x-1\right)-x^2\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(9-x^2\right)=0\)
=>\(\left(x-1\right)\left(3-x\right)\left(3+x\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\3-x=0\\3+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\\x=-3\end{matrix}\right.\)
g: \(x^3-2x^2+x-2=0\)
=>\(x^2\left(x-2\right)+\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2=0
=>x=2
h: \(\left(9-x^2\right)\left(2x+1\right)=0\)
=>(3-x)(3+x)(2x+1)=0
=>\(\left[{}\begin{matrix}3-x=0\\3+x=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
