4x^2-19x+12
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\(3x^3+19x^2+4x-12=3x^3+3x^2+16x^2+16x-12x-12=3x^2\left(x+1\right)+16x\left(x+1\right)-12\left(x+1\right)=\left(x+1\right)\left(3x^2+16x-12\right)=\left(x+1\right)\left(3x^2+18x-2x-12\right)=\left(x+1\right)\left[3x\left(x+6\right)-2\left(x+6\right)\right]=\left(x+1\right)\left(3x-2\right)\left(x+6\right)\)
3x\(^3\)-19x\(^2\)+4x+12
=3x\(^3\)-3x\(^2\)-16x\(^2\)+16x-12x+12
=3x\(^2\)(x-1)-16x(x-1)-12(x-1)
=(x-1)(3x\(^2\)-16x-12)
=(x-1)(3x\(^2\)-18x+2x-12)
=(x-1)[3x(x-6)+2(x-6)]
=(x-1)(x-6)(3x+2)
\(a,\left(12x^3y^3-3x^2y^3+4x^2y^4\right):6x^2y^3\)
\(=12x^3y^3:6x^2y^3-3x^2y^3:6x^2y^3+4x^2y^4:6x^2y^3\)
\(=2x-\frac{1}{2}+\frac{2}{3}y\)
\(b,\left(6x^3-19x^2+23x-12\right):\left(2x-3\right)\)
\(=\left(6x^3-10x^2+8x-9x^2+15x-12\right):\left(2x-3\right)\)
\(=\left(2x-3\right)\left(3x^2-5x+4\right):\left(2x-3\right)\)
\(=3x^2-5x+4\)
\(a,2x^3+5x^2+5x+3\)
\(=2x^3+3x^2+2x^2+3x+2x+3\)
\(=x^2\left(2x+3\right)+x\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(2x+3\right)\left(x^2+x+1\right)\)
d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)
\(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)
\(=x^2-2x+1\)
\(=\left(x-1\right)^2\)
4x^2-19x+12
=>4x^2-3x-16x+12
=>x(4x-3)-4(4x-3)
=>(4x-3)(x-4)
Chúc bạn làm tốt