281:
a: \(x^3+2x^2+x+2=0\)
=>\(\left(x^3+2x^2\right)+\left(x+2\right)=0\)
=>\(x^2\left(x+2\right)+\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2+1\right)=0\)
=>x+2=0
=>x=-2
b: \(x^3+2x^2-x-2=0\)
=>\(\left(x^3+2x^2\right)-\left(x+2\right)=0\)
=>\(x^2\left(x+2\right)-\left(x+2\right)=0\)
=>\(\left(x+2\right)\left(x^2-1\right)=0\)
=>\(\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\\x=-1\end{matrix}\right.\)
c: \(x^3-x^2-21x+45=0\)
=>\(x^3+5x^2-6x^2-30x+9x+45=0\)
=>\(x^2\left(x+5\right)-6x\left(x+5\right)+9\left(x+5\right)=0\)
=>\(\left(x+5\right)\left(x^2-6x+9\right)=0\)
=>\(\left(x+5\right)\left(x-3\right)^2=0\)
=>\(\left[{}\begin{matrix}x+5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
d: \(x^3+3x^2+4x+2=0\)
=>\(x^3+x^2+2x^2+2x+2x+2=0\)
=>\(x^2\left(x+1\right)+2x\left(x+1\right)+2\left(x+1\right)=0\)
=>\(\left(x+1\right)\left(x^2+2x+2\right)=0\)
=>\(\left(x+1\right)\left[\left(x+1\right)^2+1\right]=0\)
mà \(\left(x+1\right)^2+1>=1>0\forall x\)
nên x+1=0
=>x=-1
e: \(x^4+x^2+6x-8=0\)
=>\(x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)
=>\(x^3\left(x-1\right)+x^2\left(x-1\right)+2x\left(x-1\right)+8\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)
=>\(\left(x-1\right)\left[\left(x^3+8\right)+\left(x^2+2x\right)\right]=0\)
=>\(\left(x-1\right)\left[\left(x+2\right)\left(x^2-2x+4\right)+x\left(x+2\right)\right]=0\)
=>\(\left(x-1\right)\left(x+2\right)\left(x^2-x+4\right)=0\)
mà \(x^2-x+4=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>=\dfrac{15}{4}>0\forall x\)
nên (x-1)(x+2)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
g: \(\left(x^2+1\right)^2=4\left(2x-1\right)\)
=>\(x^4+2x^2+1-8x+4=0\)
=>\(x^4+2x^2-8x+5=0\)
=>\(x^4-2x^3+x^2+2x^3-4x^2+2x+5x^2-10x+5=0\)
=>\(x^2\left(x^2-2x+1\right)+2x\left(x^2-2x+1\right)+5\left(x^2-2x+1\right)=0\)
=>\(\left(x-1\right)^2\cdot\left(x^2+2x+5\right)=0\)
mà \(x^2+2x+5=\left(x+1\right)^2+4>=4>0\forall x\)
nên \(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
h: \(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)
=>\(\left(x-1+2x+3\right)\left[\left(x-1\right)^2-\left(x-1\right)\left(2x+3\right)+\left(2x+3\right)^2\right]=\left(3x+2\right)\left(9x^2-6x+4\right)\)
=>\(\left(3x+2\right)\left(x^2-2x+1-2x^2-3x+2x+3+4x^2+12x+9\right)=\left(3x+2\right)\left(9x^2-6x+4\right)\)
=>\(\left(3x+2\right)\left(3x^2+9x+13\right)-\left(3x+2\right)\left(9x^2-6x+4\right)=0\)
=>\(\left(3x+2\right)\left(3x^2+9x+13-9x^2+6x-4\right)=0\)
=>\(\left(3x+2\right)\left(-6x^2+15x+9\right)=0\)
=>\(\left[{}\begin{matrix}3x+2=0\\-6x^2+15x+9=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
i: \(6x^4-x^3-7x^2+x+1=0\)
=>\(6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1=0\)
=>\(6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(6x^3+5x^2-2x-1\right)=0\)
=>\(\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)=0\)
=>\(\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]=0\)
=>\(\left(x-1\right)\left(x+1\right)\left(6x^2-x-1\right)=0\)
=>\(\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)=0\)
=>\(\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(3x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x+1=0\\2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=\dfrac{1}{2}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
