Rút gọn (a+b)(b+c)(c-a) + (b+c)(c+a)(a-b) + (c+a)(a+b)(b-c)
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a: =b-c-a+c+1-a-b+c
=-2a+1
b: =a-b-c-b+c+a+c-b-a
=c-3b+a
c: =2(a-b-b+c-c+a)
=2(2a-2b)
=4a-4b
a) \(\left(b-c\right)-\left(a-c-1\right)-\left(a+b-c\right)\)
\(=b-c-a+c+1-a-b+c\)
\(=c-2a+1\)
b) \(\left(a-b-c\right)-\left(b-c-a\right)+\left(c-b-a\right)\)
\(=a-b-c-b+c+a+c-b-a\)
\(=a-3b+c\)
c) \(2\cdot\left(a-b\right)-2\cdot\left(b-c\right)-2\cdot\left(c-a\right)\)
\(=2\cdot\left(a-b-b+c-c+a\right)\)
\(=2\cdot\left(2a-2b\right)\)
\(=4a-4b\)
a)\(\left(a+b\right)-\left(a-b\right)+\left(a-c\right)-\left(a+c\right)=a+b-a+b+a-c-a-c\)
\(=2a-2a+2b-2c=0+2\left(b-c\right)=2\left(b-c\right)\)
b)\(\left(a+b-c\right)+\left(a-b+c\right)-\left(b+c-a\right)-\left(a-b-c\right)\)
\(=a+b-c+a-b+c-b-c+a-a+b+c\)
\(=3a-a+2b-2b+2c-2c=2a+0+0=2a\)
c)\(-\left\{-\left(a+b\right)-\left[-\left(a-b\right)-\left(a+b\right)\right]\right\}=\left\{-\left(a+b\right)-\left[b-a-a-b\right]\right\}\)
\(=-\left\{-\left(a+b\right)-b+a+a+b\right\}\)
\(=-\left\{b-a-b+2a+b\right\}\)
\(=-\left\{b-b+b+2a-a\right\}\)
\(=-\left\{b+a\right\}=-a-b\)
A=(a-b)+(a+b+c)-(a-b-c)
=a-b+a+b+c-a+b+c
=(a+a-a)+(-b+b+b)+(c+c)
= a+b+c.2
= a+b+2c
B=(a-b)-(b-c)+(c-a)-(a-b-c)
=a-b-b+c+c-a-a+b+c
=(a-a-a)+(-b-b+b)+(c+c+c)
= (-a)+ (-b) +c.3
= (-a)+(-b)+3c
C=(-a+b+c)-(a-b+c)-(a+b-c)
= (-a)+b+c-a+b-c-a-b+c
=(-a-a-a)+(b+b-b)+(c-c+c)
= (-a.3) +b+c
a + b + a + b +c + c - a - b - b + c + a - b - c -a + c + a
= a + ( - b ) + ( - c)
\(\left(a+b\right)+\left(a+b+c\right)+\left(c-a-b\right)-\left(b-c\right)+\left(a-b-c-a\right)+\left(c+a\right)\)
\(=a+b+a+b+c+c-a-b-b+c+a-b-c-a+c+a\)
\(=2a-b+3c\)
k mk nha
thank you very much
Ta có: \(\left(b-c\right)^3+\left(c-a\right)^3-\left(a-b\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(b-c+c-a\right)\left[\left(b-c\right)^2-\left(b-c\right)\left(c-a\right)+\left(c-a\right)^2\right]-\left(a-b\right)\left[1+3\left(b-c\right)\left(c-a\right)\right]\)
\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2\right)-\left(a-b\right)\left(1+3bc-3ab-3c^2+3ac\right)\)
\(=\left(b-a\right)\left(b^2-3bc+3c^2+ab-3ac+a^2+1+3bc-3ab-3c^2+3ac\right)\)
\(=\left(b-a\right)\left(b^2-2ab+a^2+1\right)\)
\(=\left(b-a\right)^3+\left(b-a\right)\)
\(=b^3-3b^2a+3ba^2-a^3+b-a\)
=(b+c)(ac-a2+bc-ab)+(b+c)(ac-bc+a2-ab)+(c+a)(a+b)(b-c)
=(b+c)(ac-a2+bc-ab+ac-bc+a2-ab)+(a+c)(a+b)(b-c)
=(b+c)(2ac-2ab)-(a+c)(a+b)(c-b)
=(b+c).2a.(c-b)-(a2+ab+ac+bc)(c-b)
=(c-b)(2ab+2ac-a2-ab-ac-bc)
=(c-b)(-a2+ab+ac-bc)=(c-b)[a(b-a)-c(b-a)]
=(c-b)(b-a)(a-c)