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câu a đặt chung x ra là xong
câu b
x^3 + 3x^2 - 7x^2 - 21x + 9x+ 27 còn lại tự làm nhé
a) x3 - 2x2 + x - xy2
= x (x2 - 2x + 1 - y2)
= x [(x2 - 2x + 1) - y2]
= x [(x - 1)2 - y2]
= x [(x - 1) + y] [(x - 1) - y]
= x (x - 1 + y) (x - 1 - y)
b) x3 - 4x2 - 12x + 27
= (x3 + 27) - (4x2 + 12x)
= (x3 + 33) - 4x (x + 3)
= (x + 3) (x2 - 3x + 32) - 4x (x + 3)
= (x + 3) [(x2 - 3x + 9) - 4x]
= (x + 3) (x2 - 3x + 9 - 4x)
= (x + 3) (x2 - 7x + 9)
#Học tôt!!!
~NTTH~
a) \(x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1\)
Nhóm các hạng tử:
\(\left(\right. x^{5} - x^{4} \left.\right) + \left(\right. - 2 x^{3} + 2 x^{2} \left.\right) + \left(\right. x - 1 \left.\right) = \left(\right. x - 1 \left.\right) \left(\right. x^{4} - 2 x^{2} + 1 \left.\right) .\)
Đặt \(t = x^{2}\) thì \(x^{4} - 2 x^{2} + 1 = \left(\right. t - 1 \left.\right)^{2} = \left(\right. x^{2} - 1 \left.\right)^{2}\).
Vậy
\(\boxed{x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1 = \left(\right. x - 1 \left.\right) \left(\right. x^{2} - 1 \left.\right)^{2} = \left(\right. x - 1 \left.\right)^{3} \left(\right. x + 1 \left.\right)^{2} .}\)
b) \(x^{3} - 5 x^{2} - 14 x\)
Lấy \(x\) chung:
\(x \left(\right. x^{2} - 5 x - 14 \left.\right) = x \left(\right. x - 7 \left.\right) \left(\right. x + 2 \left.\right) .\)
\(\boxed{x^{3} - 5 x^{2} - 14 x = x \left(\right. x - 7 \left.\right) \left(\right. x + 2 \left.\right) .}\)
c) \(2 x^{2} + 2 x y - 4 y^{2}\)
Lấy \(2\) chung: \(2 \left(\right. x^{2} + x y - 2 y^{2} \left.\right)\).
Nhân tử hóa: \(x^{2} + x y - 2 y^{2} = \left(\right. x + 2 y \left.\right) \left(\right. x - y \left.\right)\).
\(\boxed{2 x^{2} + 2 x y - 4 y^{2} = 2 \left(\right. x + 2 y \left.\right) \left(\right. x - y \left.\right) .}\)
d) \(3 x^{2} + 8 x y - 3 y^{2}\)
Thử phân tích:
\(3 x^{2} + 8 x y - 3 y^{2} = \left(\right. 3 x - y \left.\right) \left(\right. x + 3 y \left.\right) .\)
\(\boxed{3 x^{2} + 8 x y - 3 y^{2} = \left(\right. 3 x - y \left.\right) \left(\right. x + 3 y \left.\right) .}\)
e) \(x^{2} - x - x y - 2 y^{2} + 2 y\)
Gộp lại theo \(x\): \(x^{2} + x \left(\right. - 1 - y \left.\right) + \left(\right. - 2 y^{2} + 2 y \left.\right)\).
Định thức là một bình phương → nghiệm \(x = 2 y\) và \(x = 1 - y\).
Vậy
\(\boxed{x^{2} - x - x y - 2 y^{2} + 2 y = \left(\right. x - 2 y \left.\right) \left(\right. x + y - 1 \left.\right) .}\)
f) \(x^{2} + 2 y^{2} - 3 x y + x - 2 y\)
Xem như phương trình bậc hai theo \(x\): nghiệm \(x = 2 y\) và \(x = y - 1\).
Do đó
\(\boxed{x^{2} + 2 y^{2} - 3 x y + x - 2 y = \left(\right. x - 2 y \left.\right) \left(\right. x - y + 1 \left.\right) .}\)
a: \(x^5-x^4-2x^3+2x^2+x-1\)
\(=x^4\left(x-1\right)-2x^2\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^4-2x^2+1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)^2=\left(x-1\right)\cdot\left(x-1\right)^2\cdot\left(x+1\right)^2\)
\(=\left(x-1\right)^3\cdot\left(x+1\right)^2\)
b: \(x^3-5x^2-14x\)
\(=x\left(x^2-5x-14\right)\)
\(=x\left(x^2-7x+2x-14\right)\)
=x[x(x-7)+2(x-7)]
=x(x-7)(x+2)
c: \(2x^2+2xy-4y^2\)
\(=2\left(x^2+xy-2y^2\right)\)
\(=2\left(x^2+2xy-xy-2y^2\right)\)
=2[x(x+2y)-y(x+2y)]
=2(x+2y)(x-y)
d: \(3x^2+8xy-3y^2\)
\(=3x^2+9xy-xy-3y^2\)
=3x(x+3y)-y(x+3y)
=(x+3y)(3x-y)
e: \(x^2-x-xy-2y^2+2y\)
\(=\left(x^2-xy-2y^2\right)-\left(x-2y\right)\)
\(=\left(x^2-2xy+xy-2y^2\right)-\left(x-2y\right)\)
=x(x-2y)+y(x-2y)-(x-2y)
=(x-2y)(x+y-1)
f: \(x^2+2y^2-3xy+x-2y\)
\(=x^2-2xy-xy+2y^2+x-2y\)
=x(x-2y)-y(x-2y)+(x-2y)
=(x-2y)(x-y+1)
Bài giải:
a) x3 + 2x2y + xy2– 9x = x(x2 +2xy + y2 – 9)
= x[(x2 + 2xy + y2) – 9]
= x[(x + y)2 – 32]
= x(x + y – 3)(x + y + 3)
b) 2x – 2y – x2 + 2xy – y2 = (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)
c) x4 – 2x2 = x2(x2 – (√2)2) = x2(x - √2)(x + √2).
a) x3 + 2x2y + xy2– 9x = x(x2 +2xy + y2 – 9)
= x[(x2 + 2xy + y2) – 9]
= x[(x + y)2 – 32]
= x(x + y – 3)(x + y + 3)
b) 2x – 2y – x2 + 2xy – y2 = (2x – 2y) – (x2 – 2xy + y2)
= 2(x – y) – (x – y)2
= (x – y)[2 – (x – y)]
= (x – y)(2 – x + y)
c) x4 – 2x2 = x2(x2 – (√2)2) = x2(x - √2)(x + √2).
a) x2 – 4 + (x – 2)2
= (x2 – 22) + (x – 2)2 = (x – 2)(x + 2) + (x – 2)2
= (x – 2) [(x + 2) + (x – 2)]
= (x – 2)(x + 2 + x – 2)
= 2x(x – 2)
b) x3 – 2x2 + x – xy2
= x(x2 – 2x + 1 – y2) = x[(x2 – 2x + 1) – y2]
= x[(x – 1)2 – y2]
= x[(x – 1) + y] [(x – 1) – y]
= x(x – 1 + y)(x – 1 – y)
c) x3 – 4x2 – 12x + 27
= (x3 + 27) – 4x(x + 3)
= (x + 3)(x2 – 3x + 9) – 4x(x + 3)
= (x + 3)(x2 – 3x + 9 – 4x)
= (x + 3)(x2 – 7x + 9)
a. 2x2-xy
= x(2x-y)
b. x2-xy-x+y
= (x2-xy)-(x+y)
=x(x-y)-(x-y)
=(x-y)(x-1)
a) = (x3 +33) -4x(x+3)
= (x+3)(x2 -3x+9-4x)
= (x+3)(x2 - 7x +9)
x3 -2x2 +x- xy2
= x ( x2 - 2x + 1 - y2)
= x\(\left[\left(x-1\right)^2-y^2\right]\)
= x ( x- 1- y) ( x - 1 + y )
X3-2x2+x-xy2
=(x3+x)-(2x2-xy2)
=x(x2+1)-x(2x-y2)
=x(x2-2x+1-y2)
=x[(x-1)2-y]
=x(x-1-y)(x-1+y)
Chúc bạn làm tốt@"
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1+y\right)\left(x-1-y\right)\)
\(a^3-a^2x-ay+xy\)
\(=a^2\left(a-x\right)-y\left(a-x\right)\)
\(=\left(a-x\right)\left(a^2-y\right)\)
\(4x^2-y^2+4x+1\)
\(=\left(4x^2+4x+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2=\left(2x-y+1\right)\left(2x+y+1\right)\)
\(x^3-x+y^3-y\)
\(=\left(x^3+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
a)a3 - a2x - ay +xy
=(a3 - a2x) - (ay - xy)
=a2(a-x) - y(a-x)
=(a-x).(a2 - y)
x3 – 2x2 + x – xy2
(Có nhân tử chung x)
= x(x2 – 2x + 1 – y2)
(Có x2 – 2x + 1 là hằng đẳng thức).
= x[(x – 1)2 – y2]
(Xuất hiện hằng đẳng thức (3))
= x(x – 1 + y)(x – 1 – y)