a)3x²+7x-6

=3x²-2x+9x-6

=x(3x-2)+3(3x-2)

=(x+3)(3x-2)

b)8x²-2x-3

=8x²-6x+4x-3

=2x(4x-3)+(4x-3)

=(2x+1)(4x-3)

c)6x²-15x+6

=3(2x²-5x+2)

=3(2x²-x-4x+2)

=3[x(2x-1)-2(2x-1)]

=3(x-2)(2x-1)

d)10x²+7x-6

=10x²+12x-5x-6

=2x(5x+6)-(5x+6)

=(2x-1)(5x+6)

a) \(\left(x^2-7x+12\right).\left(x^2-15x+56\right)-60\)

\(=\left(x-3\right)\left(x-4\right)\left(x-7\right)\left(x-8\right)-60\)

b) \(x^4+2000x^2+1999x+2000\)

\(=\left(x^2-x+2000\right)\left(x^2+x+1\right)\)

\(=\left(x^2+x-1\right)^2+1999\left(x^2+x+1\right)+1\)

hình như Thành lộn đề rồi đấy ạ, ý mình là phân tích thành nhân tử á

b) \(x^4+2000x^2+1999x+2000\)

\(=x^4-x+2000x^2+2000x+2000\)

\(=x\left(x^3-1\right)+2000\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2000\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x-2000\right)\)

x³ + 3x² + 4x² + 12x + 3x + 9

= x²(x + 3) + 4x(x + 3) + 3(x + 3)

= (x + 3)(x² +4x + 3)

=(x +3)(x² + x + 3x + 3)

=(x + 3)²(x + 1)

\(x^3-2x^2-5x+6\)

\(=\left(x^3-4x^2+3x\right)+\left(2x^2-8x+6\right)\)

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

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

1. \(x^3-2x-5x+6\)

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

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

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

2. \(x^3-7x^2+15x-9\)

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

\(\Leftrightarrow x^2\left(x-1\right)-6x\left(x-1\right)+9\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-6x+9\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x\left(x-3\right)-3\left(x-3\right)\right)\)

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

\(x^3-7x^2+15x-9\)

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

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

\(=\left(x-1\right)\left(x^2-6x+3\right)\)

   6x^2 + 15x - 36

= 6x^2 + 24x - 9x - 36

= (6x^2 + 24x) - (9x + 36)

= 6x(x + 4) - 9(x+4)

= (6x - 9) (X + 4)

= 3(2x - 3)(x + 4)

\(6x^2+15-36\)

\(=3\left(2x^2+5x-12\right)\)

\(=3\left(2x^2+8x-3x-12\right)\)

\(=3\left[2x\left(x+4\right)-3\left(x+4\right)\right]\)

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

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

=6x^2-12x-3x+6

=6x(x-2)-3(x-2)

=(6x-3)(x-2)

ừ hè quên mất 12 với 3