Question

chung to rang:

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

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

Answer 1

\(\left(a+b\right)-\left(b-a\right)+c\)

\(=\left(a+b\right)-b+a+c\)

\(=2a+c\)

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

\(-2b=\left[\left(-a\right)+\left(-b\right)-\left(-c\right)\right]+\left(a-b-c\right)\)

\(-2b=\left[\left(-a\right)+\left(-b\right)-\left(-c\right)\right]+a-\left(b+c\right)\)

(-a) + a = 0 nên ta có

\(\left[\left(-b\right)-\left(-c\right)\right]-\left(b+c\right)=\left[\left(-b\right)+c\right]-\left(b+c\right)\)

\(=-2b\left(đpcm\right)\)

Answer 2

co dung ko ban