\(\left(x+1\right)^2\)\(-...">
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\(\left(x+1\right)^2-\left(x-1\right)^2\)

\(\Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)\)

\(\Leftrightarrow2.2x=4x\)

p/s tham khảo nha

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

\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-\left(a-b\right)\)

\(\Leftrightarrow\left(a-b\right)\left(a+b-1\right)\)

p/s tham khảo

15 tháng 9 2017

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

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

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

21 tháng 10 2019

a) -3x^2+x+4

=-3x^2-3x+4x+4

=-3x.(x+1)+4.(x+1)

=(x+1).(4-3x)

21 tháng 10 2019

b) \(x^3-3x^2+2\)

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

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

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

c) \(x^4y^4+64\)

\(=x^4y^4+16x^2+64-16x^2\)

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

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

d) \(x^8+x^7+1\)

\(=x^8+x^7+x^6-x^6+1\)

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

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

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

\(=\left(x^2+x+1\right)\left[x^6-x^4-x+x^3-1\right]\)

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

Bài 2:

a)A= \(6x^2\)\(-11x+3\)

<=>A=\(6x^2\)\(-2x-9x+3\)

<=>A=(\(6x^2\)\(-2x\))-\(\left(9x-3\right)\)

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

<=>A=\(2x\left(3x-1\right)+3\left(3x-1\right)\)

=>A=(3x-1)(2x+3)

1 tháng 10 2017

a)\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\text{[}\left(b^3-c^3\right)+\left(a^3-b^3\right)\text{]}+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)

\(=\left(a-b\right)\left(b^3-c^3\right)-\left(b-c\right)\left(a^3-b^3\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(bc+c^2-a^2-ab\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)

4 tháng 8 2017

a, \(x^4-x^3-x^3+x^2-x^2+x+x-1\)\(1\)

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

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

b, \(\left(ab-1\right)^2+\left(a+b\right)^2\)

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

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

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

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

c,\(x^4+2x^3+2x^2+2x+1\)

=\(x^4+x^3+x^3+x^2+x^2+x+x+1\)

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

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

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

24 tháng 10 2018

khong biet

24 tháng 10 2018

tui đếch bt vì tui mới hk lớp 5  thôi à