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a) \(12x^3+8x^2-3x-2=4x^2\left(3x+2\right)-\left(3x+2\right)\)
\(=\left(3x+2\right)\left(4x^2-1\right)=\left(3x+2\right)\left(2x-1\right)\left(2x+1\right)\)
b) \(18x^3+27x^2-2x-3=9x^2\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(9x^2-1\right)=\left(2x+3\right)\left(3x-1\right)\left(3x+1\right)\)
c) \(8x^3+4x^2-34x+15=4x^2\left(2x-3\right)+8x\left(2x-3\right)-5\left(2x-3\right)\)
\(=\left(2x-3\right)\left(4x^2+8x-5\right)=\left(2x-3\right)\left(2x-1\right)\left(2x+5\right)\)
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b) \(x^3-4x^2-12x+27=\left(x^3+27\right)-\left(4x^2+12x\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-3x+9-4x\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
d) \(x^{16}-1=\left(x^4-1\right)\left(x^4+1\right)=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\)
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\(x^3+2x^2+x-4xy^2\)
\(=x\left(x^2+2x+1\right)-4xy^2\)
\(=x\left(x+1\right)^2-4xy^2\)
\(=x\left(\left(x+1\right)^2-4y^2\right)\)
\(=x\left(\left(x+1-2y\right)\left(x+1+2y\right)\right)\)
\(\text{x3+2x2+x−4xy2 =x(x2+2x+1)−4xy2 =x(x+1)2−4xy2 =x((x+1)2−4y2) =x((x+1−2y)(x+1+2y))}\)
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a) = (x-3)(x+3) +(x-3((x-3)
= (x-3)(x+3+x-3)
= 2x(x-3)
làm cho 1 câu thui
![](https://rs.olm.vn/images/avt/0.png?1311)
b)3x^2-18x+27=3x^2-9x-9x+27=3x*(x-3)-9*(x-3)=(x-3)*(3x-9)=(x-3)*3*(x-3)=3*(x-3)^2
c)x^3-4x^2-12x+27=(x+3)*(x^2-3x+9-4)=(x+3)*(x^2-3x+5)
d)27x^3-1/27=(3x-1/3)*(9x^2-x+1/9) (hang dt)
con a) voi e) mk chiu
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( 4x+1) (12x-1) (3x+2) (x+1) -4
=(4x+1)(3x+2)(12x-1)(x+1)-4
=(12x2+11x+2)(12x2+11x-1)-4
Đặt t=12x2+11x+2 ta được:
t.(t-3)-4
=t2-3t-4
=t2+t-4t-4
=t.(t+1)-4.(t+1)
=(t+1)(t-4)
thay t=12x2+11x+2 ta được:
(12x2+11x+3)(12x2+11x-2)
Vậy ( 4x+1) (12x-1) (3x+2) (x+1) -4=(12x2+11x+3)(12x2+11x-2)
b) (x2+2x)2+9x2+18x+20
=(x2+2x)2+9.(x2+2x)+20
Đặt y=x2+2x ta được:
y2+9y+20
=y2+4y+5y+20
=y.(y+4)+5.(y+4)
=(y+4)(y+5)
thay y=x2+2x ta được:
(x2+2x+4)(x2+2x+5)
Vậy (x2+2x)2+9x2+18x+20=(x2+2x+4)(x2+2x+5)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\)
\(4x^2-y^2+2x+y\)
\(=\left(4x^2-y^2\right)+\left(2x+y\right)\)
\(=\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)\)
\(=\left(2x+y\right)\left(2x-y+1\right)\)
\(b)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x-9\right)\)
\(=\left(x-3\right)\left(x^2+5-9\right)\)
\(c)\)
\(12x^3+4x^2-27x-9\)
\(=\left(12x^3+4x^2\right)-\left(27x+9\right)\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)[\left(2x\right)^2-3^2]\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(d)\)
\(16x^2+4x-y^2+y^2\)
\(=16x^2+4x\)
\(4x\left(4x+1\right)\)
a.2x2-12x2+18x
=>2x(x2-6x+9)
=>2x(x-3)2
b.(x2+x)2+4x2+4x-12
=>x4+2x3+x2+4x2+4x-12
=>x4+2x3+5x2+4x-12
a) \(2x^3-12x^2+18x=2x\left(x^2+6x+9\right)=2x\left(x+3\right)^2\)
b) \(\left(x^2+x\right)^2+4x^2+4x-12\)
\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
\(=\left(x^2+x\right)^2+6\left(x^2+x\right)-2\left(x^2+x\right)-12\)
\(=\left(x^2+x\right)\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+6\right)\)
\(=\left(x^2-x+2x-2\right)\left(x^2+x+6\right)\)
\(=\left[x\left(x-1\right)+2\left(x-1\right)\right]\left(x^2+x+6\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)