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e) Ta có: x4−2x3+2x−1x4−2x3+2x−1
=(x4−1)−2x(x2−1)=(x4−1)−2x(x2−1)
=(x2+1)(x−1)(x+1)−2x(x−1)(x+1)=(x2+1)(x−1)(x+1)−2x(x−1)(x+1)
=(x−1)(x+1)⋅(x2−2x+1)=(x−1)(x+1)⋅(x2−2x+1)
=(x+1)⋅(x−1)3=(x+1)⋅(x−1)3
h) Ta có: 3x2−3y2−2(x−y)23x2−3y2−2(x−y)2
=3(x2−y2)−2(x−y)2=3(x2−y2)−2(x−y)2
=3(x−y)(x+y)−2(x−y)2=3(x−y)(x+y)−2(x−y)2
=(x−y)(3x+3y−2x+2y)=(x−y)(3x+3y−2x+2y)
=(x−y)(x+5y)=(x−y)(x+5y)
b) (1 + 2x)(1- 2x) - x(x+2)(x-2)
= (1- 4x2) - x(x2 - 4)
= 1 - 4x2- x3- 4x
= (1 - x3) + (4x - 4x2)
= (1- x) (1 + x + x2) + 4x(1 -x)
= (1-x)(1+5x + x2)
(x-1)(x+1)(x+2)
=(x^2-1)(x+2)
=x^3+2x^2-x-2
[X-1/2] [X+1/2] [4X-1]
=\(\left(x^2-\frac{1}{4}\right)\left(4x-1\right)\)
=\(4x^3-x^2-x+\frac{1}{4}\)
1/2X2Y2 [2X+Y] [2X-Y]
=\(\frac{1}{2}x^2y^2\left(4x^2-y^2\right)\)
=\(2x^2y^2-\frac{1}{2}x^2y^4\)
a) 16x2(x - y)2 - 10y(y - x)3
= 16x2(y - x)2 - 10y(y - x)3
= 2(y - x)2[8x2 - 5y(y - x)]
= 2(y - x)2(8x2 + 5xy - 5y2)
b) a2 -b2 + 4ab - 9 (sai đề)
a) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\\ =\left(x+2\right)^2-y.\left(x+2\right)\)
\(=\left(x+2\right).\left(x+2-y\right)\)
b) \(2x^2-5x-3\)
\(=2x^2+x-6x-3\)
\(=\left(2x^2+x\right)-\left(6x+3\right)=x\left(2x+1\right)-3\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x-3\right)\)
c)\(\)
c);d);e) tạm thời tớ chưa nghĩ ra-.-"
tham khả tạm 2 câu ạ, chúc học tốt'.'
a, \(x^3-2x^2+3x-6=x\left(x^2+3\right)-2\left(x^2+3\right)=\left(x-2\right)\left(x^2+3\right)\)
b, \(x^2+2x+1-4y^2=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)