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a.(a+b)-(b+a)+c=
=a+b-b+a+c
=2a+c (đpcm)
Vậy (a+b)-(b-a)+c=2a+c
b.-(a+b-c)+(a-b-c)=
=-a-b+c+a-b-c
=-b-b
=-2b (đpcm)
Vậy -2b=-(a+b+c)+(a-b-c)
NHớ tick cho mình nha!!!!!!!!!
\(\left(a-b+c\right)-\left(a+c\right)\)
\(=a-b+c-a-c\)
\(=\left(a-a\right)+\left(c-c\right)-b\)
\(=0+0-b\)
\(=0-b\)
\(=-b\)
1) (a - b + c) - (a + c)
= a - b + c - a - c
= (a - a) - b + (c - c)
= 0 - b + 0 = -b
2) (a + b) - (b - a) + c
= a + b - b + a + c
= (a + a) + (b - b) + c
= 2a + 0 + c = 2a + c
3) -(a + b - c) + (a - b - c)
= -a - b + c + a - b - c
= (-a + a) - (b + b) + (c - c)
= 0 - 2b + 0 = -2b
4) a(b + c) - a(b + d)
= ab + ac - ab - ad
= (ab - ab) + a(c - d)
= 0 + a(c - d) = a(c - d)
5) tự lm
(a-b+c)-(a+c)=a+c-a-c-b=-b
(a+b)-(b-a)+c=a+b-b+a+c=2a+c
-(a+b-c)+(a-b-c)=a-b-c-a-b+c=-2b
a(b+c)-a(b+d)=a(b+c-b-d)=a(c-d)
a(b-c)+a(d+c)=a(b-c+d+c)=a(b+d)
bạn ấn vào đúng 0 sẽ ra kết quả, mình làm bài này rồi dễ lắm
1 , a - ( a - b - c ) - ( b - c -a ) - ( c - b -a )
= a - a + b + c - b + c + a - c + b + a
= (a-a+a) + (b-b+b) + (c-c+c)
= a+b+c
2 , - ( a + b + c ) - ( b - c -a ) + ( 1 - a - b ) - ( c - 3b )
= -a - b - c - b + c + a + 1 - a - b - c + 3b
= (a+a-a) - (b+b+b) + (c-c+c) + 3b
= a - 3b + c + 3b
= a+c + (3b - 3b)
= a+c + 0
= a+c
3 , ( b - c - 6 ) - ( 7 - a + b ) + c
= b - c - 6 - 7 + a - b + c
= (b-b) + (c-c) - (6+7) + a
= 0 + 0 - 13 + a
= -13 + a
4 , - ( 3b - 2a - c ) - ( a - b - c ) - ( a - 2b -+ 2c )
= -3b + 2a + c - a + b + c - a + 2b - 2c
= -3b + (2b + b) + (c + c) - (a+a) +2a - 2c
= -3b + 3b + 2c - 2a + 2a - 2c
= (3b - 3b) + (2c - 2c) + (2a + 2a)
= 0 + 0 + 0
= 0
chỉ bt lm đến đây thoy
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''';l';.;';p''ơ'a, (a-b+c)-(a+c)= a-b+c-a-c=a-a+c-c-b=-b
b, -(a+b-c)+(a-b-c)=-a-b+c+a-b-c=-a+a-b-b+c-c=-2b
1) a - ( a - b - c ) - ( b - c - a ) - ( c - b - a )
= a - a + b + c - b + c + a - c + b + a
= 2a + b + c
2) - ( a + b + c ) - ( b - c - a ) + ( 1 - a - b ) - ( c - 3b )
= -a - b - c - b + c + a + 1 - a - b - c + 3b
= 1 - a - c
1,a-(a-b-c)-(b-c-a)-(c-b-a)
=a-a+b+c-b+c+a-c+b+a
=2a+b+c
2,-(a+b+c)-(b-c-a)+(1-a-b)-(c-3b)
=-a-b-c-b+c+a+1-a-b-c+3b
=1-a-c
3,(b-c-6)-(7-a+b)+c
=b-c-6-7+a-b+c
=a-13
4,-(3b-2a-c)-(a-b-c)-(a-2b+2c)
=-3b+2a+c-a+b+c-a+2b-2c
=0
5,(4a-3b+2c)-(4b-3c-2a)-(4c-3a+2b)+(a-b)-c
=4a-3b+2c-4b+3c+2a-4c+3a-2b+a-b-c
=(4a+2a+3a+a)-(3b+4b+2b+b)+(2c+3c-4c-c)
=10a-10b+0
=10.(a-b)
6,
2a-{a-b[a-b-(a+b+c)+2b]-c-b}
=2a-{a-b[a-b-a-b-c+2b]-c-b}
=2a-a-bc+c+b
=a-bc+c+b
=(a+b)-b(c-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
\(\left[\left(-b\right)-\left(-c\right)\right]-\left(b+c\right)=\left[\left(-b\right)+c\right]-\left(b+c\right)\)
\(=-2b\left(đpcm\right)\)
co dung ko ban