\(x^4-x^3+6x^2-x+a=x^2\left(x^2-x+5\right)+x^2-x+a\)

Do \(x^2\left(x^2-x+5\right)\) chia hết \(x^2-x+5\)

\(\Rightarrow x^2-x+a\) chia hết \(x^2-x+5\)

\(\Rightarrow a=5\)

\(a,\Leftrightarrow x^3-8-x^3-2x=12\Leftrightarrow-2x=20\Leftrightarrow x=-10\\ b,\Leftrightarrow x^2-6x+9-x^2+4=16\Leftrightarrow=-6x=3\Leftrightarrow x=-\dfrac{1}{2}\\ c,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-6\right)+9\left(x-6\right)=0\\ \Leftrightarrow\left(x^2+9\right)\left(x-6\right)=0\\ \Leftrightarrow x=6\left(x^2+9>0\right)\)

A

A. (x-2)³ = x³ - 6x² +12x - 8 (hằng đẳng thức)

a)
=(x-2)³
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
 

a) \(3x^2-6xy=3x\left(x-2y\right)\)

b) \(x^3-6x^2+9x=x\left(x^2-6x+9\right)=x\left(x-3\right)^2\)

c) \(=x\left(x-2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x-3\right)\)

d) \(=2x\left(3x-5\right)-3\left(3x-5\right)=\left(3x-5\right)\left(2x-3\right)\)

\(a,=3x\left(x-2y\right)\\ b,=x\left(x-3\right)^2\\ c,Sửa:x^2-2xy-3x+6y=x\left(x-2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x-3\right)\\ d,=\left(3x-5\right)\left(2x-3\right)\)

\(\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

a) Ta có: \(x^3+x^2+x+1=0\)

\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)

mà \(x^2+1>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

b) Ta có: \(x^3-6x^2+11x-6=0\) 

\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)

\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

Vậy: S={1;2;3}

c) Ta có: \(x^3-x^2-21x+45=0\)

\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)

\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2\cdot\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: S={3;-5}

d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)

\(\Leftrightarrow x^3\left(x-2\right)+4x^2\cdot\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+4x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+3x^2+x^2+4x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+3\right)+\left(x+1\right)\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)=0\)

mà \(x^2+x+1>0\forall x\)

nên (x-2)(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy: S={2;-3}

a: \(C=x^3-3x^2+3x+2023\)

\(C=x^3-3x^2+3x-1+2024\)

\(=\left(x-1\right)^3+2024\)

Khi x=101 thì \(C=\left(101-1\right)^3+2024\)

\(=100^3+2024\)

\(=1000000+2024=1002024\)

b: \(D=x^3-6x^2+12x-100\)

\(=x^3-6x^2+12x-8-92\)

\(=\left(x-2\right)^3-92\)

Khi x=-98 thì \(D=\left(-98-2\right)^3-92\)

\(=-100^3-92\)

\(=-1000000-92=-1000092\)

\(a,=5xy\left(2x-y+3z\right)\\ b,=x^2\left(x-1\right)-4\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\\ c,=x\left(x^2-6x+9\right)=x\left(x-3\right)^2\)