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a) \(x^{12}-3x^6+1\)
\(=\left(x^6\right)^2-2\cdot x^6\cdot1+1^2-x^6\)
\(=\left(x^6-1\right)^2-\left(x^3\right)^2\)
\(=\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=\left(x^2\right)^2+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)
a, \(x^4+6x^3+7x^2-6x+1\)
\(=x^4-2x^2+1+6x^3+9x^2+6x\)
\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)
\(=\left(x^2-1+3x\right)^2\)
b, \(x^4-7x^3+14x^2-7x+1\)
\(=x^4+2x^2+1+7x^3+12x^2-7x\)
\(=\left(x^2+1\right)^2-7x\left(x^2+1\right)+12^2\)
\(=\left(x^2-1+3x\right)^2\)
c, \(12x^2-11x-36\)
\(=12x^2-27x+16x-36\)
\(=3x\left(4x-9\right)+4\left(4x-9\right)\)
\(=\left(4x-9\right)\left(3x+4\right)\)
x3 + 7x - 6=x2 . x + 7x - 22 + 2 = (x2 - 22) + (x+7x)+2=(x-2) . (x+2) + 8x + 2
x3 - 5x + 8x - 4=x2 . x -5x + 8x -22 = (x2 - 22) . (x -5x + 8x )=(x-2) . (x+2) . 4x
x3 - 9x2 + 6x + 16=x2 . x - 9x2 + 6x + 16 = (x2 - 9x2) . (x+6x) + 16=(x-9x) . (x+9x) . 7x + 16
k mk nha
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)
\(x^3-5x^2-14x\)
\(=x^3+2x^2-7x^2-14x\)
\(=x^2\left(x+2\right)-7x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-7x\right)\)
\(=x\left(x+2\right)\left(x-7\right)\)
\(x^3-7x-6\)
\(=x^3+x^2-x^2-x-6x-6\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)\)
\(=\left(x+1\right)\left(x^2+2x-3x-6\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x-3\right)\)
\(x^3-19x-30\)
\(=x^3-5x^2+5x^2-25x+6x-30\)
\(=x^2\left(x-5\right)+5x\left(x-5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+5x+6\right)\)
\(=\left(x-5\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x-5\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x+3\right)\left(x+2\right)\)
1)\(8x^6-\frac{1}{125}y^3=\left(2x^2\right)^3-\left(\frac{1}{5}y\right)^3\)
Bạn tự lm tiếp.AD HĐT số (7)
2)\(\left(x+4\right)^3-64=\left(x+4\right)^3-4^3\)
AD HĐT số (7).Tự lm tiếp
3)\(x^6+1=\left(x^2\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
4)\(x^9+1=\left(x^3\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
5,\(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2\)
AD HĐT số (3).Tự lm tiếp
6)\(x^3+6x^2+12x+8=\left(x+2\right)^3\)
AD HĐT số (4)
7)\(x^3-15x^2+75x-125=\left(x-5\right)^3\)
AD HĐT số (5)
8)\(27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.\left(3a\right)^2.2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
AD HĐT số (5)
a) \(x^3-7x-6=x^3-x^2+x^2-7x-6=x^2\left(x-1\right)+x^2-x-6x+6\)
\(=x^2\left(x-1\right)+\left(x\left(x-1\right)-6\left(x-1\right)\right)\)
\(=\left(x-1\right)\left(x^2+x-6\right)=\left(x-1\right)\left(x^2-2x+3x-6\right)\)
\(\left(x-1\right)\left(x\left(x-2\right)+3\left(x-2\right)\right)=\left(x-1\right)\left(x-2\right)\left(x+3\right)\)
b)\(x^3-x^2-14x+24=x^3-3x^2+2x^2-6x-8x+24\)
\(=x^2\left(x-3\right)+2x\left(x-3\right)-8\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+2x-8\right)=\left(x-3\right)\left(x^2-2x+4x-8\right)\)
\(=\left(x-3\right)\left(x\left(x-2\right)+4\left(x-2\right)\right)=\left(x-3\right)\left(x-2\right)\left(x+4\right)\)
CÓ CHỖ NÀO KO HIỂU GỬI THƯ HỎI MIK , MIK NÓI CHO !!~ HOK TỐT ~
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).