Thu gọn : C = a ( b + c ) - b ( a - c ) - c ( a + b ) + 3 ac
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\(a,A=\left(2a+b+3c\right)-\left(a-b+c\right)\)
\(=2a+b+3c-a+b-c\)
\(=a+2b-2c\)
\(b,B=\left(a+b-c\right)-\left(-2a+b-c\right)-\left(-a-b-2c\right)\)
\(=a+b-c+2a-b+c+a+b+2c\)
\(=4a+b+2c\)
\(c,C=\left(a-2b-c\right)-\left(-2a+b-c\right)-\left(-a-b-2c\right)\)
\(=a-2b-c+2a-b+c+a+b+2c\)
\(=4a-2b+2c\)
1: =a-b+c-a-c=-b
2: =a+b-b+a+c=2a+c
3: =-a-b+c+a-b-c=-2b
4: =ab+ac-ab-ad-ac+ad=0
1) \(a\left(b-c-d\right)-a\left(b+c-d\right)\)
\(=ab-ac-ad-ab-ac+ad\)
\(=-2ac\)
2) \(\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)\)
\(=ac+ad+bc+bd-ab-ac-bd-cd\)
\(=ad+bc-ab-cd\)
3) \(\left(a+b\right)\left(c-d\right)-\left(a-b\right)\left(c+d\right)\)
\(=ac-ad+bc-bd-ac-ad+bc+bd\)
\(=-2ad+2bc\)
\(=-2\left(ad-bc\right)\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)^3+c^3+3c.\left(a+b\right).\left(a+b+c\right)\right]-a^3-b^3-c^3\)
\(=\left[a^3+b^3+3ab.\left(a+b\right)+c^3+3c.\left(a+b\right)\right]-a^3-b^3-c^3\)
\(=3ab.\left(a+b\right)+3c.\left(a+b\right)\left(a+b+c\right)=3.\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng :
Đặt \(\left\{{}\begin{matrix}a+b-c=x\\a-b+c=y\\-a+b+c=z\end{matrix}\right.\) \(\Rightarrow x+y=z=a+b+c\)
Khi đó biểu thức trở thành :
\(\left(x+y+z\right)^3-x^3-y^3-z^3=3.\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(=3.2a.2b.2c=24abc\)
=ab+ac-ab+bc-ac-bc+3ac
=3ac
Ta có: C = a (b + c) - b (a - c) - c (a + b) + 3 ac
= ab + ac - ba + bc - ca - cb + 3 ac
= (ab - ba) + (ac - ca) + (bc - cb) + 3 ac
= 0 + 0 + 0 + 3 ac
= 3 ac