@Quang
1) Ta có:
(a-b+c) - (a+c) = a - b + c - a - c = (a - a) - b + (c - c) = -b
Vậy (a-b+c) - (a+c) = -b (dpcm)
2) Ta có :
(a+b) - (b-a) + c = a + b - b + a + c = (a + a )+ (b -b) + c = 2a + c
Vậy (a+b) - (b-a) + c = 2a + c (dpcm)
3)Ta có:
-(a+b-c) + (a-b-c) = -a - b + c + a - b - c = (-a + a) - (b+b) + (c -c) = 0 - 2b + 0 = -2b
Vậy -(a+b-c) + (a-b-c) = -2b (dpcm)
4)Ta có:
a(b+c) - a(b+d) = a [(b+c) - (b+d)] = a.(b+c - b -d ) = a.(b -b + c - d) = a.(c-d)
Vậy a(b+c) - a(b+d) = a.(c-d) (dpcm)
5)Ta có :
a.(b-c) + a.(d+c) = a [(b-c) + (d+c)] = a.(b-c+d+c) = a(b+d -c+c) = a(b+d)
Vậy a.(b-c) + a.(d+c) = a(b+d) (dpcm)
@Mong tus cho hay nhất
