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a ) a + b = -1 => a = -1 - b
b + c = 1 => c = 1 - b
Thay vào a + c = 6.
Ta được : -1 - b + 1 - b = 6
=> -2b = 6
=> b = -3
=> a = -1 - - 3 = 2
=> c = 1 - - 3 = 4
b ) ab = - 35
=> \(a=\dfrac{-35}{b}\)
bc = 7
\(\Rightarrow c=\dfrac{7}{b}\)
Thay vào abc = 35, ta được :
\(\dfrac{-245}{b}=35\Leftrightarrow35b=-245\Rightarrow b=-7\)
=> \(a=-\dfrac{35}{-7}=\dfrac{35}{7}\)
=> \(c=\dfrac{7}{-7}=-1\)
c ) Đặt a + b + c = -6 (1)
b + c + d = -9 (2)
c + d + a = -8 (3)
d + a + b = -7 (4)
Cứ thay từ từ rồi sẽ ra .

a, \(\left(a-b\right)+\left(c-d\right)=\left(a+c\right)-\left(b+d\right)\)
\(a-b+c-d=a+c-b-d\)
\(\Rightarrow VT=VP\left(đpcm\right)\)
b, \(\left(a-b\right)-\left(c-d\right)=\left(a+d\right)-\left(b+c\right)\)
\(a-b-c+d=a+d-b-c\)
\(\Rightarrow VT=VP\left(đpcm\right)\)
c, \(a-\left(b-c\right)=\left(a-b\right)+c=\left(a+c\right)-b\)
\(a-b+c=a-b+c=a+c-b\)
\(\Rightarrowđpcm\)
d, \(\left(a-b\right)-\left(b+c\right)+\left(c-a\right)-\left(a-b-c\right)=-\left(a+b-c\right)\)
\(a-b-b-c+c-a-a+b+c=-a-b+c\)
\(-a-b+c=-a-b+c\)
\(\Rightarrow VT=VP\left(đpcm\right)\)
e, \(-\left(-a+b+c\right)+\left(b+c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\)
\(a-b-c+b+c-1=b-c+6-7+a-b+c\)
\(a-1=-1+a\Rightarrow a-1=a+\left(-1\right)\Rightarrow a-1=a-1\)
\(\Rightarrow VT=VP\left(đpcm\right)\)

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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\(a\)) \(Ta\) \(có\)\(:\) \(\left(a-b\right)-\left(c-d\right)=a-b-c+d\)
\(=a+\left(-b\right)+\left(-c\right)+d\)
\(=\left(a+d\right)+\left[\left(-b\right)+\left(-c\right)\right]\)
\(=\left(a+d\right)+\left[-\left(b+c\right)\right]\)
\(=\left(a+d\right)-\left(b+c\right)\)

Ta có
\(\left(a-b\right)+\left(c-d\right)=a-b+c-d=\left(a+c\right)-\left(b+d\right)\)
b
\(\left(a-b\right)-\left(c-d\right)=a-b-c+d=\left(a+d\right)-\left(b+c\right)\)
c,
\(-\left(-a+b+c\right)+\left(b+c-1\right)=a-b-c+b+c-1=\left(b-c+6\right)-\left(7-a+b\right)+c\)Nếu thấy bài làm của mình đúng thì tick nha ban.Nhân dịp đầu xuân năm mới mình chúc bạn vui vẻ mạnh khoẻ nha.

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)
a) (b - c + 6) - (7 - a + b) + c
= b - c + 6 - 7 + a - b + c
= (b-b)+(-c+c)+(6-7)+a
= -1 + a = a - 1
b) -(a-b-c) + (-c+b+a) - (a+b)
= -a+b+c-c+b+a-a-b
= (-a+a-a) + (b+b-b)+(c-c)
= -a + b + 0
= -a + b = b - a