Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^3-2x^2-x\)
\(=x\cdot x^2-x\cdot2x-x\cdot1\)
\(=x\left(x^2-2x-1\right)\)
x4+2.x3-13.x2-14x+24
=x3.(x+2)-13x2+12x-26x+24
=x3.(x+2)-x.(13x-12)-2.(13x-12)
=x3.(x+2)-(13x-12)(x+2)
=(x+2)(x3-13x+12)
=(x+2)(x3-x-12x+12)
=(x+2)[x.(x2-1)-12.(x-1)]
=(x+2)[x.(x-1)(x+1)-12.(x-1)]
=(x+2)(x-1)[x.(x+1)-12]
=(x+2)(x-1)(x2+x-12)
=(x+2)(x-1)(x2-3x+4x-12)
=(x+2)(x-1)[x.(x-3)+4.(x-3)]
=(x+2)(x-1)(x-3)(x+4)
\(x^3-x^2-14x+24\)
\(=x^3-2x^2+x^2-2x-12x+24\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right)\left(x^2+4x-3x-12\right)\)
\(=\left(x-2\right)\left[x\left(x+4\right)-3\left(x+4\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
Ta có:\(x^3-x^2-14x+24=\left(x^3-2x^2\right)+\left(x^2-2x\right)-\left(12x-24\right)\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right)\left(x^2-3x+4x-12\right)\)
\(=\left(x-2\right)\left[x\left(x-3\right)+4\left(x-3\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
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) \(x^{5} - x^{4} - 2 x^{3} + 2 x^{2} + x - 1\)
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)
đa thức trên không thể phân tích dù tớ đã vào Cốc Cốc mà cốc cốc cũng bó tay
\(=\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left(x+6\right)\left(x+4\right)\left(x+2\right)\left(x+8\right)+16\)
\(=\left(x^2+10x+24\right)\left(x^2+10x+16\right)+16\)
\(=\left(t+8\right)t+16=\left(t+4\right)^2=\left(x^2+10x+20\right)^2\)
Ta có: \(x+x^2+x^3+x^4+x^5+x^6+\cdots+x^{13}+x^{14}+x^{15}\)
\(=\left(x+x^2+x^3\right)+\left(x^4+x^5+x^6\right)+\cdots+\left(x^{13}+x^{14}+x^{15}\right)\)
\(=x\left(1+x+x^2\right)+x^4\left(1+x+x^2\right)+\cdots+x^{13}\left(1+x+x^2\right)\)
\(=\left(x^2+x+1\right)\left(x+x^4+\cdots+x^{13}\right)\)
\(=x\left(x^2+x+1\right)\left(1+x^3+x^6+x^9+x^{12}\right)\)
x(1+x*1+...+x*14)