\(x^2+3ab\left(2-3ab\right)-10xy-1+25y^2\)

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18 tháng 8 2017

b) \(a^5+a^4+a^3+a^2+a+1\)

\(=a^4\left(a+1\right)+a^2\left(a+1\right)+a+1\)

\(=\left(a^4+a^2+1\right)\left(a+1\right)\)

c) \(x^3-1+5x^2-5+3x-3\)

\(=\left(x-1\right)\left(x^2+x+1\right)+5\left(x^2-1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left[x^2+x+1+5\left(x+1\right)+3\right]\)

\(=\left(x-1\right)\left(x^2+6x+9\right)=\left(x-1\right)\left(x+3\right)^2\)

18 tháng 8 2017

a) \(x^2+3ab\left(2-3ab\right)-10xy-1+25y^2\)

\(=\left(x^2-10xy+25y^2\right)+6ab-9a^2b^2-1\)

\(=\left(x-5y\right)^2-\left(9a^2b^2-6ab+1\right)\)

\(=\left(x-5y\right)^2-\left(3ab-1\right)^2\)

\(=\left(x-5y-3ab+1\right)\left(x-5y+3ab-1\right)\)

1 tháng 11 2016

Đây, bản full đây thím, tớ thực sự đã kiên nhẫn lắm đấy ...

a)\(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)=4\left(x^2-y^2-2x+2ay-a^2+1\right)\)

\(=4\left[\left(x^2-2x+1\right)-\left(a^2-2ay+y^2\right)\right]\)

\(=4\left[\left(x-1\right)^2-\left(a-y\right)^2\right]\)

\(=4\left(x-1-a+y\right)\left(x-1+a-y\right)\)

b)\(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1\right)-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)

\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)

c)\(x^3-1+5x^2-5+3x-3=\left(x-1\right)\left(x^2+x+1\right)+5\left(x^2-1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)+5\left(x-1\right)\left(x+1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)\left(5x+5\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1+5x+5+3\right)\)

\(=\left(x-1\right)\left(x^2+6x+9\right)\)

\(=\left(x-1\right)\left(x+3\right)^2\)

d)\(a^5+a^4+a^3+a^2+a+1=a^4\left(a+1\right)+a^2\left(a+1\right)+\left(a+1\right)\)

\(=\left(a+1\right)\left(a^4+a^2+1\right)\)

\(=\left(a+1\right)\left(a^4+2a^2+1-a^2\right)\)

\(=\left(a+1\right)\left[\left(a^2+1\right)^2-a^2\right]\)

\(=\left(a+1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)\)

e)\(x^3-3x^2+3x-1-y^3=\left(x-1\right)^3-y^3\)

\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)

\(=\left(x-1-y\right)\left(x^2-2x+1+xy-y+y^2\right)\)

f)\(5x^3-3x^2y-45xy^2+27y^3=5x\left(x^2-9y^2\right)-3y\left(x^2-9y^2\right)\)

\(=\left(x^2-9y^2\right)\left(5x-3y\right)\)

\(=\left(x-3y\right)\left(x+3y\right)\left(5x-3y\right)\)

g)\(3x^2\left(a-b+c\right)+36xy\left(a-b+c\right)+108y^2\left(a-b+c\right)\)

\(=\left(a-b+c\right)\left(3x^2+36xy+108y^2\right)\)

\(=3\left(a-b+c\right)\left(x^2+12xy+36y^2\right)\)

\(=3\left(a-b+c\right)\left(x+6y\right)^2\)

1 tháng 11 2016

a/ \(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\)

\(=\left(4x^2-8x+4\right)-\left(4y^2-8ay+4a^2\right)\)

\(=\left(2x-2\right)^2-\left(2y-2a\right)^2=\left(2x-2+2y-2a\right)\left(2x-2-2y+2a\right)\)

b/ \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)=\left(x+y-1\right)\left(x^2+y^2+2xy+x+y+1\right)-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)\)

Giải giúp bạn 2 bài tiêu biểu thôi nha

3 tháng 7 2018

a)   \(x^2+2x+1=\left(x+1\right)^2\)

b)   \(9x^2+y^2+6xy=\left(3x+y\right)^2\)

c)   \(25a^2+4b^2-20ab=\left(5a-2b\right)^2\)

d)   \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)

e)   \(\left(2x+3y\right)^3+2\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)

f) mk chỉnh lại đề nha:

 \(2xy^2+x^2y^4+1=\left(xy^2+1\right)^2\)

g)  \(x^2+6xy+9y^2=\left(x+3y\right)^2\)

h)  \(x^2-10xy+25y^2=\left(x-5y\right)^2\)

3 tháng 7 2018

cảm ơn bn nha!

9 tháng 8 2018

mk ghi đáp án, còn lại bạn tự biến đổi

a) \(2x^3-x^2+5x+3=\left(2x+1\right)\left(x^2-x+3\right)\)

b) \(x^3+5x^2+8x+4=\left(x+1\right)\left(x+2\right)^2\)

c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

d) \(4x^4+1=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)

e) \(x^4-7x^3+14x^2-7x+1=\left(x^2-4x+1\right)\left(x^2-3x+1\right)\)

9 tháng 8 2018

mk làm chi tiết theo yêu của của người hỏi đề:

a) \(2x^3-x^2+5x+3\)

\(=\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\)

\(=2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\)

\(=\left(2x+1\right)\left(x^2-x+3\right)\)

b)  \(x^3+5x^2+8x+4\)

\(=\left(x^3+4x^2+4x\right)+\left(x^2+4x+4\right)\)

\(=x\left(x^2+4x+4\right)+\left(x^2+4x+4\right)\)

\(=\left(x+1\right)\left(x^2+4x+4\right)\)

\(=\left(x+1\right)\left(x+2\right)^2\)

24 tháng 10 2018

khong biet

24 tháng 10 2018

tui đếch bt vì tui mới hk lớp 5  thôi à

15 tháng 9 2017

a, \(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)

      \(=x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

       \(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)

6 tháng 9 2020

a) \(\left(x+y\right)^3-x^3-y^3\)

\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=\left(x+y\right)\left[\left(x+y\right)^2-x^2+xy-y^2\right]\)

\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)

\(=3xy\left(x+y\right)\)

b) \(x^2+y^2+2xy+yz+xz\)

\(=\left(x^2+2xy+y^2\right)+\left(yz+xz\right)\)

\(=\left(x+y\right)^2+z\left(x+y\right)\)

\(=\left(x+y\right)\left(x+y+z\right)\)

c) \(x^2-10xy-1+25y^2\)

\(=\left(x^2-10xy+25y^2\right)-1\)

\(=\left(x-5y\right)^2-1\)

\(=\left(x-5y-1\right)\left(x-5y+1\right)\)

d) \(ax^2-ax+bx^2-bx+a+b\)

\(=(ax^2+bx^2)-(ax+bx)+(a+b)\)

\(=x^2(a+b)-x(a+b)+(a+b)\)

\(=(a+b)(x^2-x+1)\)

e)\(x^2-2y+3xz+x-2y+3z\)

\(=(x^2+x)-(2xy+2y)+(3xz+3z)\)

\(=x(x+1)-2y(x-1)+3z(x+1)\)

\(=(x+1)(x-2y+3z)\)

f) \(xyz-xy-yz-xz+x+y+z-1\)

\(=(xyz-xy)-(yz-y)-(xz-x)+(z-1)\)

\(=xy(z-1)-y(z-1)-x(z-1)+(z-1)\)

\(=(z-1)(xy-y-x+1)\)

\(=(z-1)[y(x-1)-(x-1)]\)

\(=(z-1)(x-1)(y-1)\)

_Học tốt_