\(a,A=\left(2a+b+3c\right)-\left(a-b+c\right)\)

\(=2a+b+3c-a+b-c\)

\(=a+2b-2c\)

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

\(=a+b-c+2a-b+c+a+b+2c\)

\(=4a+b+2c\)

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

\(=a-2b-c+2a-b+c+a+b+2c\)

\(=4a-2b+2c\)

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

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nhanh lên

mk sắp đi học rùi

ai nhanh nhất mk kick

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 :

\(\left(x+y+z\right)^3-x^3-y^3-z^3=3.\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(=3.2a.2b.2c=24abc\)