\(1.\left(a-b+c\right)-\left(a+c\right)=-b\)
Ta có: \(a-b+c-a-c\)
\(=a-a-b+c-c\)
\(=-b\left(ĐPCM\right)\)
\(2.\left(a+b\right)-\left(b-a\right)+c=2a+c\)
Ta có: \(a+b-b+a+c\)
\(=a+a+b-b+c\)
\(=2a+c\left(ĐPCM\right)\)
1.(a-b+c)-(a+c)
=a-b+c-a-c
=-b (đổi vị trsi là ra)
=> đpcm
Lm hơi tắt, sr..
\(3.-\left(a+b-c\right)+\left(a-b-c\right)=-2b\)
Ta có: \(-a-b+c+a-b-c\)
\(=-a+a-b-b+c-c\)
\(=-2b\)
1/ Ta có :
\(\left(a-b+c\right)-\left(a+c\right)\)
\(=a-b+c-a-c\)
\(=-b\left(đpcm\right)\)
2/ Ta có :
\(\left(a+b\right)-\left(b-a\right)+c\)
\(=a+b-b+c+c\)
\(=2a+c\left(đpcm\right)\)
3/ Ta có :
\(-\left(a+b-c\right)+\left(a-b-c\right)\)
\(=-a-b+c+a-b-c\)
\(=-2b\left(đpcm\right)\)
2.\(\left(a+b\right)-\left(b-a\right)+c\)
=\(a+b-b+a+c\)
=\(2a+c\)
=> Đpcm
3.-(a+b-c)+(a-b-c)
=-a-b+c+a-b-c
=-2b
=> Đpcm