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1: =a-b+c-a-c=-b
2: =a+b-b+a+c=2a+c
3: =-a-b+c+a-b-c=-2b
4: =ab+ac-ab-ad-ac+ad=0
a, \(\left(a+b+c+d\right)-\left(a-b+c-d\right)\)
\(=a+b+c+d-a+b-c+d=2b+2d\)
\(=2.\left(b+d\right)\)
b, \(\left(-a+b-c+d\right)+\left(a+d\right)+\left(-b+c\right)\)
\(=-a+b-c+d+a+d-b+c\)
\(=2d\)
c, \(-\left(a-b-d\right)+\left(b-c+d\right)-\left(-c+b-d\right)\)
\(=-a+b+d+b-c+d+c-b+d\)
\(=-a+b+3d\)
Chúc bạn học tốt!!!
a,(a+b+c+d)-(a-b+c-d) = a + b + c + d - a + b - c + d = (a - a) + (b + b) + (c - c) + (d + d) = 2b + 2d = 2.(b + d)
b,(-a+b-c+d)+(a+d)+(-b+c) = -a + b - c + d + a + d - b + c = (-a + a) + (b - b) + ( -c + c) + (d + d) = 2d
c,-(a-b-d)+(b-c+d)-(-c+b-d)= -a + b + d + b - c + d + c - b + d = -a + (b + b - b) + (-c + c) + ( d + d + d) = -a + b + 3d
Bài 1 :
a, Rút gọn :
A = ( - a - b + c ) - ( - a - b - c )
= - a - b + c + a + b + c
= 2c
b, Thay c = - 2 vào biểu thức A = 2c
Ta được : A = 2 x ( - 2 ) = - 4
Bài 3 : Ta có : A + B = a + b - 5 + ( - b - c + 1 )
= a + b - 5 - b - c + 1
= a - c - 4
C - D = b - c - 4 - ( b - a )
= b - c - 4 - b + a
= a - c - 4
=> A + B = C - D ( đpcm )
a) \(\left(a+b-c\right)-\left(b-c+d\right)\)
\(=a+b-c-b+c-d\)
\(=a+b-b-c+c-d\)
\(=a-d\)
b) \(-\left(a-b+c\right)+\left(a-b+d\right)\)
\(=\left(-a\right)+b-c+a+b-d\)
\(=\left(-a\right)+a+b+b-c-d\)
\(=2b-c-d\)
c) \(\left(a+b\right)-\left(-a+b-c\right)\)
\(=a+b+a-b+c\)
\(=a+a+b-b+c\)
\(=2a+c\)
d) \(-\left(a-b-c\right)+\left(a-b-c\right)\)
\(=-a+b+c+a+b+c\)
\(=-a+a+b+b+c+c\)
\(=2b+2c\)
(chắc hơi sai... > . < ...)
1) -a(-a+c-d)-(c-a+d)
=ad-ac+a2-c+a-d
2) -(a+b)-(c+d)+(a-b-c-d)
=-a-b-c-d+a-b-c-d
=-2b-2c-2d
3) a(b-c-a)-a(b+c-d)
=-ac+ab-a2+ad-ac-ab
=ad-2ac-a2
bài 1:
a) ta có: 3x + 5 = (3(x+1)+2)\(⋮\)(x+1)
vì (3(x+1)\(⋮\)(x+1) nên 2 \(⋮\)(x+1) => (x+1) \(\in\)Ư(2) => (x+1) \(\in\)\(\xi\)-2;-1;1;2 \(\xi\)=> x \(\in\)\(\xi\)-3; -2; 0; 1 \(\xi\)
vậy, x= -3; -2; 0; 1
a, -( -a + c - d) - ( c - d + d) = a - c + d - c + d - d = a + d
b, - ( a+b-c+d) + (a-b-c-d) = -a -b+c-d + a-b-c-d = -2b + (-2c)= -2(b+c)
1) \(a\left(b-c-d\right)-a\left(b+c-d\right)\)
\(=ab-ac-ad-ab-ac+ad\)
\(=-2ac\)
2) \(\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)\)
\(=ac+ad+bc+bd-ab-ac-bd-cd\)
\(=ad+bc-ab-cd\)
3) \(\left(a+b\right)\left(c-d\right)-\left(a-b\right)\left(c+d\right)\)
\(=ac-ad+bc-bd-ac-ad+bc+bd\)
\(=-2ad+2bc\)
\(=-2\left(ad-bc\right)\)
A = ( a + b ) - ( d - b ) - ( c + d )
A = a + b - d + b - c - d
Thay a = -2 , b = 3 vào biểu thức trên ta được :
- 2 + 3 - d + 3 - c - d
= - 2 + ( 3 + 3 ) - ( d - d ) - c = - 2 + 6 - 0 - c = 4 - c
\(1.\left(a-b\right)+\left(b-a-c\right)=a-b+b-a-c=-c\)
\(2.\left(a-b\right)-\left(a-b+c\right)=a-b-a+b-c=-c\)
\(3.\left(-a+c-d\right)-\left(c-a+d\right)=-a+c-d-c+a-d=-2d\)\
\(4.\left(b+d\right)+\left(-a+b+d\right)=b+d-a+b+d=-a+2b+2d\)
\(5.\left(a+b+c\right)-\left(a+b-5\right)=a+b+c-a-b+5=c+5\)
\(6.\left(a-b-c\right)-\left(-a-b-c\right)-\left(-a+b-c\right)-\left(c-a-b\right)+\left(-3a-c\right)\)
\(=a-b-c+a+b+c+a-b+c-c+a+b-3a-c\)
\(=\left(a+a+a+a-3a\right)+\left(-b+b-b\right)\)
Xl mình làm lại câu 6
\(\left(a-b-c\right)-\left(-a-b-c\right)-\left(-a+b-c\right)-\left(c-a-b\right)+\left(-3a+c\right)\)
\(=a-b-c+a+b+c+a-b+c-c+a+b-3a+c=a+c\)