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a) a3+b3+a2c+b2c-abc
= (a+b)(a2-ab+b2)+c(a2+b2)-abc
=(a+b) [ (a+b)2-3ab]+c.[(a+b)2-2ab]-abc
=(a+b)(a+b)2-3ab(a+b)+c(a+b)2-3abc
=(a+b)2(a+b+c)-3ab(a+b+c)
=(a+b)2.0-3ab.0
=0
b) ax+ay+2x+2y+4
=a(x+y)+2(x+y)+4
=(x+y)(a+2)+4
=(a-2)(a+2)+4
=a2-4+4
=a2
c) A=1+x+x2+...+x49=>Ax=x+x2+x3+...+x50
- A=1+x+x2+...+x49
---> Ax-A=x50-1
d)(a+b)(a+c)+(c+a)(c+b)
=a2+ac+ab+bc+c2+bc+ac+ab
=a2+c2+2ac+2ab+2bc
=2b2+2bc+2ac+2ab
=2b(b+c)+2a(b+c)
=2b(b+c)(b+a)
1 ) Ta có :
\(a+b-c=0\Leftrightarrow a+b=c\Leftrightarrow\left(a+b\right)^3=c^3\)
\(\Rightarrow a^3+b^3-c^3=a^3+b^3-\left(a+b\right)^3\)
\(\Rightarrow a^3+b^3-c^3=a^3+b^3-3a^2b-3b^2a-b^3\)
\(\Rightarrow a^3+b^3-c^3=-3a^2b-3b^2a\)
\(\Rightarrow a^3+b^3-c^3=-3ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3-c^3=-3abc\left(đpcm\right)\)
2 ) Ta có :
\(a-b+c=0\Leftrightarrow c=b-a\Leftrightarrow c^3=\left(b-a\right)^3\)
\(\Rightarrow a^3-b^3+c^3=a^3-b^3+\left(b-a\right)^3\)
\(\Rightarrow a^3-b^3+c^3=a^3-b^3+b^3-3a^2b+3b^2a-a^3\)
\(\Rightarrow a^3-b^3+c^3=-3a^2b+3b^2a\)
\(\Rightarrow a^3-b^3+c^3=-3ab\left(a-b\right)\)
\(\Rightarrow a^3-b^3+c^3=3ab\left(b-a\right)\)
\(\Rightarrow a^3-b^3+c^3=3abc\left(đpcm\right)\)
1 ) Bổ sung dấu \(\Rightarrow\) thứ 2 :
\(\Rightarrow...=a^3+b^3-a^3-3a^2b-3b^2a-b^3\)
Bài 2:
\(A=\left(2ac-a^2-c^2+b^2\right)\left(2ac+a^2+c^2-b^2\right)\)
\(=\left[b^2-\left(a-c\right)^2\right]\left[\left(a+c\right)^2-b^2\right]\)
\(=\left(b-a+c\right)\left(b+a-c\right)\left(a+c-b\right)\left(a+c+b\right)\)>0
Ta có:
\(A=a^3+a^2c-abc+b^2c+b^3=0\Rightarrow\left(a^3+b^3\right)+\left(a^2c+b^2c-abc\right)=0\)
\(\Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)=0\Rightarrow\left(a+b+c\right)\left(a^2-ab+b^2\right)=0\)
Mà theo giả thiết thì \(a+b+c=0\Rightarrow A=0\)
P/s: Lười ghi nên đổi thành A nhé ;)
Ta có :
\(a^3+a^2c-abc+b^2c+b^3=0\)
\(\Leftrightarrow\left(a^3+b^3\right)+\left(a^2c-abc+b^2c\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)=0\)
\(\Leftrightarrow\left(a^2-ab+b^2\right)\left(a+b+c\right)=0\) ( Luôn đúng vì \(a+b+c=0\) )
Wish you study well !!
Ta có: \(a^3+a^2c-abc+b^2c+b^3\)
\(=a^3+b^3+c\left(a^2+b^2-ab\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)
\(=\left(a^2-ab+b^2\right)\left(a+b+c\right)\)
\(=0\)(vì a+b+c=0)
Vậy \(a^3+a^2c-abc+b^2c+b^3=0\left(\text{đ}pcm\right)\)
Ta có: a+b+c=0 nên a= -(b+c) ; b= -(a+c) ; c= -(b+c). Khi đó:
a3 + a2c -abc + b2c +b3 = a2 (a+b) + b2 (b+c) -abc = -(a2b +ab2) -abc = -ab(a+b) -abc =abc -abc = 0 (đpcm)