Ta có : x3 + 2x2 + x
= x3 + x2 + x2 + x
= x2(x + 1) + x(x + 1)
= (x2 + x) (x + 1)
= x(x + 1)(x + 1)
Ta có : xy + y2 - x - y
= y(x + y) - (x + y)
= (x + y)(y - 1)
c) a3 +b3+c3-3abc
= (a+b)3-3a2b-3ab2+c3-3abc
=[(a+b)3+c3 ]-3ab(a+b)-3abc
=[(a+b)+c][(a+b)2-(a+b)c+c2]-3ab(a+b+c)
=(a+b+c){[(a+b)2-(a+b)c+c2]-3ab}
=(a+b+c)(a2+b2+2ab-ac-bc+c-3ab)
=(a+b+c)(a2+b2+c2-ac-bc-ab)
\(a,\)\(x^3+2x^2+x\)
\(=x.\left(x^2+2x+1\right)\)
\(=x.\left(x+1\right)^2\)
\(b,\)\(xy+y^2-x-y\)
\(=\left(xy-x\right)+\left(y^2-y\right)\)
\(=x.\left(y-1\right)+y.\left(y-1\right)\)
\(=\left(y-1\right).\left(x+y\right)\)
a)\(x^3+2x^2+x=x\left(x^2+2x+1\right)\)
\(=x\left(x+1\right)^2\)
b)\(xy+y^2-x-y=xy-x+y^2-y\)
\(=x\left(y-1\right)+y\left(y-1\right)\)
\(=\left(y-1\right)\left(x+y\right)\)
c)\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Câu này là 1 hằng đẳng thức
