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\(\Leftrightarrow x^3-2x^2+x^2-2x-6x+12\)

\(\Leftrightarrow x^2\left(x-2\right)+x\left(x-2\right)-6\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x-6\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+3x-2x-6\right)\)

\(\Leftrightarrow\left(x-2\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-2\right)\)

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

T I C K ủng hộ nha 

_________________CHÚC BẠN HỌC TỐT ___________________

Tùng làm đúng nhưng bài này cỉ một vế nên sử dụng dấu bằng nếu thi chuyển cấp, mà bài 1 vế sử dụng dấu tương đương sẽ mất điểm cà bài. còn nếu 1 dấu thì 0,25 đấy nên sừ dụng dấu bằng nha

\(x^2-8x+12=\left(x^2-6x\right)-\left(2x-12\right)=x\left(x-6\right)-2\left(x-6\right)=\left(x-2\right)\left(x-6\right)\)

=(x-2)(x-6)

x³-x²-8x+12

=x³-2x²+x²-2x-6x+12

=(x³-2x²)+(x²-2x)-(6x-12)

=x²(x-2)+x(x-2)-6(x-2)

=(x-2)(x²+x-6)

chúc bạn học tốt

= \(x^4-2x^3-6x^3+12x^2-x^2+2x+6x-12\)

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

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

= \(\left(x-2\right)\left(x^2\left(x-6\right)-\left(x-6\right)\right)\)

= \(\left(x-2\right)\left(x-6\right)\left(x-1\right)\left(x+1\right)\)

x⁴ - 8x³ + 11x² + 8x - 12

= (x³ - 7x² + 4x + 12)(x - 1)

= (x³ - 8x + 12)(x + 1)(x - 1)

= (x - 6)(x - 2)(x + 1)(x - 1)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(=\left(x-3\right)\left(8x^3-16x^2\right)=8x^2\left(x-2\right)\left(x-3\right)\)

\(8x^3\left(x-3\right)+16x^2\left(3-x\right)\)

\(=8x^3\left(x-3\right)-16x^2\left(x-3\right)\)

\(=8x^2\left(x-3\right)\left(x-2\right)\)