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28 tháng 12 2019

\(a^3-a^2+2\)

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

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

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

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

20 tháng 9 2020

 .\(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-2abc-a^3-b^3-c^3\)

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

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

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

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

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

26 tháng 1 2019

a) (a-b)(b-c)(a-c).

b) (a-b)(b-c)(a - c)(a + b + c).

12 tháng 8 2018

a) ( 3 x   -   2 y ) 3 .        b) ( x   -   1 ) ( x   +   3 ) 2 .

\(a\left(b^2+c^2\right)+b\left(a^2+c^2\right)+c\left(a^2+b^2\right)-2abc-a^3-b^3-c^3\)

\(=c\left(a-b\right)^2+\left[ab^2+ac^2+a^2b+bc^2-a^3-b^3-c^3\right]\)

\(=c\left(a-b\right)^2+c^2\left(a+b-c\right)+ab^2+a^2b-a^3-b^3\)

\(=c\left(a-b\right)^2+c^2\left(a+b-c\right)-\left(a^3-a^2b\right)+\left(ab^2-b^3\right)\)

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

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

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

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

24 tháng 4 2019

a3 ( c - b2 ) + b3 ( a - c2 ) + c3 ( b - a2 ) + abc ( abc - 1 )

= a3c - a3b2 + b3a - b3c2 + c3b - c3a2 + a2b2c2 - abc

= a2b2c2 - b3c2 - ( a2c3 - bc3 ) - ( a3b2 - ab3 ) + ( a3c - abc )

= b2c2 . ( a2 - b ) - c3 ( a2 - b ) - ab2 ( a2 - b ) + ac ( a2 - b ) 

= ( a2 - b ) ( b2c2 - c3 - ab2 + ac )

= ( a2 - b ) ( b2 - c ) ( c2 - a )

20 tháng 7 2021

a) x2-4y2-x++2y

= x2-(2y)2-x+2y

= (x-2y)(x+2y)-(x-2y)

=(x-2y)(x+2y-1)

18 tháng 8 2023

\(a,27x^3-54x^2y+36xy^2-8y^3\)

\(=\left(3x\right)^3-3.\left(3x\right)^2.2y+3.3x.\left(2y\right)^2-\left(2y\right)^3\)

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

\(b,x^3-1+5x^2-5+3x-3\)

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

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

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

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

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

\(c,a^5+a^4+a^3+a^2+a+1\)

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

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

18 tháng 8 2023

\(27x^3-54x^2y+36xy^2-8y^3\)

\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)

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

______________________

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

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

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

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

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

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

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

________________

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

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

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

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

AH
Akai Haruma
Giáo viên
23 tháng 7 2021

Lời giải:

a. $a^4+a^3+a^2+a=(a^4+a^3)+(a^2+a)$

$=a^3(a+1)+a(a+1)=(a+1)(a^3+a)=a(a+1)(a^2+1)$
b. $3xy^2+5y-3x^2y+(-5x)=(3xy^2-3x^2y)+(5y-5x)$

$=3xy(y-x)+5(y-x)=(y-x)(3xy+5)$

c. $xy-z+y-xz=(xy+y)-(z+xz)=y(x+1)-z(x+1)=(x+1)(y-z)$

d.

$x^2-bx+ax-ab=(a^2+ax)-(bx+ab)=a(a+x)-b(a+x)=(a+x)(a-b)$

a(b3 - c3) + b(c- a3) + c(a- b3)

= a(b3 - c) + b( c3 - b3 + b3 - a3) + c(a3 - b3)

= a(b3 - c3) + b(c3 - b3) + b(b3 - a3) + c(a3 - b3)

= a(b3 - c3) - b(b3 - c3) - [b(a3 - b3) - c(a3- b3)]

= (b3 - c3)(a - b) - (a3- b3)(b - c)

= (b - c)(b2 + bc + c2)(a - b) - (a - b)(a2 + ab + b2)(b - c)

= (b - c)(a - b)(b2 + bc + c2 - a2 + ab - b2)

= (b - c)(a - b) [ (c2  - a2) + (bc - ab) ]

= (b - c)(a - b) [ (c - a)(c + a) + b(c - a) ]

= (b - c)(a -b) [ (c - a)(c + a + b) ]

= (a- b)(b - c)(c - a)(a + b + c)