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19 tháng 10 2018
a3 + b3 + c3 - 3abc
= (a3 + 3a2b + 3ab2 + b3 ) + c3 - 3abc - 3a2b - 3ab2
=[(a+b)3 + c3 ]- (3abc+3a2b+3ab2)
=(a+b+c)[(a+b)2 - (a+b)c + c2 ] - 3ab(c+a+b)
=(a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)
=(a+b+c)(a2+2ab+b2-ac-bc+c2-3ab)
=(a+b+c)(a2+b2+c2-ab-bc-ca)
19 tháng 8 2016
Bài 1 :
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
Bài 2 : Ta có : \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3-3abc=-c^3\) ( Vì \(a+b=-c\) )
\(\Rightarrow a^3+b^3+c^3=3abc\)
19 tháng 8 2016
Bài 1:
x2 +4x-y2+4
=(x2+4x+4)-y2
=(x+2)2-y2
=(x-y+2)(x+y+2)
Bài 2:
a3+b3+c3 = 3abc
=>a3+b3+c3-3abc=0
=>[(a+b)3+c3]-3ab(a+b)-3abc=0
=>(a+b+c)[(a+b)2-(a+b)c+c2]-3ab(a+b+c)=0
=>(a+b+c)(a2+b2+c2-ac-bc-ab)=0
Từ a+b+c=0
=>0*(a2+b2+c2-ac-bc-ab)=0 (luôn đúng)
BD
0
HA
2
25 tháng 8 2019
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(a^3-c^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left[\left(a^3-b^3\right)+\left(b^3-c^3\right)\right]+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)
\(=\left(b^3-c^3\right)\left(a-b\right)-\left(a^3-b^3\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(b^2+bc+c^2\right)\left(a-b\right)-\left(a-b\right)\left(a^2+ab+b^2\right)\left(b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(b^2+bc+c^2\right)-\left(a^2+ab+b^2\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(bc+c^2-a^2-ab\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[b\left(c-a\right)+\left(c-a\right)\left(c+a\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
TN
28 tháng 8 2019
a(b3−c3)+b(c3−a3)+c(a3−b3)
=a(b3−c3)−b(a3−c3)+c(a3−b3)
=a(b3−c3)−b[(a3−b3)+(b3−c3)]+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)
=(a−b)(b−c)[(b2+bc+c2)−(a2+ab+b2)]
=(a−b)(b−c)(bc+c2−a2−ab)
=(a−b)(b−c)[b(c−a)+(c−a)(c+a)]
=(a−b)(b−c)(c−a)(a+b+c)
HA
0
NN
0
VA
1
12 tháng 7 2016
\(B=x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-xz\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
TH
1
29 tháng 9 2019
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3a^2b+3ab^2+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+2abc+b^2c+bc^2\right)\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+abc+abc+b^2c+bc^2\right)\)
\(=3\left[ab\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+c^2+ac+bc\right)\)
\(=3\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
a^3 + b^3 +c^3 - 3abc
= ( a + b )^3 - 3ab ( a + b ) + c^3 - 3abc
= [ ( a + b )^3 + c^3 ] + [ -3ab ( a + b ) - 3abc ]
= ( a + b + c ) * [ ( a + b )^2 - c*( a + b ) + c^2 ] - 3ab ( a + b + c )
= ( a + b + c ) * ( a^2 + 2ab + b^2 - ac - bc + c^2 - 3ab )
= (a + b + c) * ( a^2 + b^2 + c^2 - ab - ac - bc )