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\(x^4+6x^3+12x^2+8x\)
\(=x\left(x^3+6x^2+12x+8\right)\)
\(=x\left(x+2\right)^3\)
8x + 12x2 + 6x3 + x4
= x4 + 6x3 + 12x2 + 8x
= x(x3 + 6x2 + 12x + 8)
= x ( x + 2 ) 3
đưa về dạng hằng đẳng thức thứ 4 lập phương của 1 tổng
\(x^3+6x^2+12x+8\)
\(=\left(x+2\right)^3\)
\(x^3+6x^2+12x+8=x^3+2.3x^2+2.3^2x+2^3=\left(x+2\right)^3\)
xong ròi k1 mình nha bn thanks
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, \(12x^2-11x-36\)
\(=12x^2-27x+16x-36\)
\(=3x\left(4x-9\right)+4\left(4x-9\right)\)
\(=\left(4x-9\right)\left(3x+4\right)\)
Mình ko thêm bớt hạng tử nhé.
\(8x^3-3x+6x^2-1\)
\(=\left(8x^3-1\right)+\left(6x^2-3x\right)\)
\(=\left(2x-1\right)\left(4x^2+2x+1\right)+3x\left(2x-1\right)\)
\(=\left(2x-1\right)\left[\left(4x^2+2x+1\right)+3x\right]\)
\(=\left(2x-1\right)\left(4x^2+5x+1\right)\)
\(=\left(2x-1\right)\left[4x\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(2x-1\right)\left(x+1\right)\left(4x+1\right)\)
\(8x^3-3x+6x^2-1=\left(8x^3-12x^2+6x-1\right)+\left(18x^2-9x\right)\)
\(=\left(\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\right)+\left(18x^2-9x\right)\)
\(=\left(2x-1\right)^3+9x\left(2x-1\right)=\left(2x-1\right)\left(\left(2x-1\right)^2+9x\right)\)
\(=\left(2x-1\right)\left(4x^2-4x+1+9x\right)=\left(2x-1\right)\left(4x^2+5x+1\right)\)
\(3x^4+6x^3-7x^2+8x-10\)
\(=\left(3x^4-3x^3\right)+\left(9x^3-9x^2\right)+\left(2x^2-2x\right)+\left(10x-10\right)\)
\(=\left(x-1\right)\left(3x^3+9x^2+2x+10\right)\)
\(a.=x^3+3x^2y+3x^2y+9xy^2+3xy^2+9y^3\)
\(=x^2\left(x+3y\right)+3xy\left(x+3y\right)+3y^2\left(x+3y\right)\)
\(=\left(x+3y\right)\left(x^2+3xy+3y^2\right).\)
\(b.=9x^3+3x^2y+9x^2y+3xy^2+3xy^2+y^3\)
\(=3x^2\left(3x+y\right)+3xy\left(3x+y\right)+y^2\left(3x+y\right)\)
\(=\left(3x^2+3xy+y^2\right)\left(3x+y\right)\).
1) \(\left(3x^2-3y^2\right)-\left(12x-12y\right)\)
\(=3xy\left(x-y\right)-12\left(x-y\right)\)
\(=\left(3xy-12\right)\left(x-y\right)\)
2) \(4x^3+4xy^2+8x^2y-16x\)
\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)
\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)
Ta có : 3x2 - 3y2 - 12x + 12y
= (3x2 - 3y2) - (12x - 12y)
= 3(x2 - y2) - 12(x - y)
= 3(x - y)(x + y) - 4.3.(x - y)
= 3(x - y)(x + y - 4)
(2x-1)^3