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A+B=a+b-5+(-b-c+1)=a+b-5-b-c+1=a-c-4 (1)
C-D=b-c-4-(b-a)=b-c-4-b+a=a-c-4 (2)
từ (1) và (2) suy ra A+B=C-D
a,M=(a+b+c-d)+(a-b+c-d)=a+b+c-d+a-b+c-d=2a+2c-2d=2(a+c-d)
b,-(a+b-c)+(a-b-c)
-a-b+c+a-b+c=-2b (ĐPCM)
hok tút
b) -(a + b - c) + (a - b - c)
= -a - b + c + a - b - c
= (-a + a) - (b + b) + (c - c)
= 0 - 2b + 0
= -2b
a, Bạn ơi sao đề này lạ z bạn???? Ko tin thì mik làm thử cho nè
(a+b)+(c-d)-(a+c)=-(b+d)
<=> a+b+c-d-a-c=-b-d
<=>(a-a)+(c-c)+(b-d)=-b-d
<=>0+0+b-d=-b-d-d
=>b-d=-b
Đó lạ ko!!!
b, Đề bài này thì mik nghĩ đúng:
(a-b)-(c-d)+(b+c)=a+d
<=>a-b-c+d+b+c=a+d
<=>(-b+b)+(-c+c)+a+d=a+d
<=>0+0+a+d=a+d
=>a+d=a+d(dpcm)
1, a(b+c)-b(a-c)=(a+b)c
\(ab+ac-ba+bc=\left(a+b\right)c\)
\(a.\left(b-b\right)+\left(a+b\right).c=\left(a+b\right)c\)
\(a.0+\left(a+b\right)c=\left(a+b\right)c\)
\(\left(a+b\right)c=\left(a+b\right)c\)
\(\Rightarrowđpcm\)
2, a(b-c)-a(b+d)=-a(c+d)
\(ab-ac-ab-ad=a.\left(c+d\right)\)
\(a.\left(b-c-b-d\right)=a\left(-c-d\right)\)
\(a.\left(-c-d\right)=a.\left(-c-d\right)\)
\(\Rightarrowđpcm\)
3, (a+b)(c+d)-(a+d)(b+c)=(a-c)(d-b)
=ac+ad+bc+bd-ab-ac-bd-dc
=ad-ab+bc-dc
=(ad-ab)+(bc-dc)
=a(d-b)+c(b-d)
=a(d-b)-c(d-b)
=(a-c)(d-b) =VP.
\(\Rightarrowđpcm\)
học tốt
1,a.(b+c)-b.(a-c)
=a.b+a.c-(b.a-b.c)
=a.b+a.c-b.a+b.c
=(a.b-b.a)+(a.c+b.c)
=0+c.(a+b)=c.(a+b)
2)a.(b-c)-a.(b+d)
=a.b-a.c-(a.b+a.d)
=a.b-a.c-a.b-a.d
=(a.b-a.b)-a.c-a.d
=0-a.c-a.d
=-a.c-a.d
=-a.c+(-a.d)
=-a.(c+d)
3)(a+b).(c+d)-(a+d).(b+c)
=a.c+a.d+a.c+a.d-(a.b+a.c+d.b+d.c)
=a.c+a.d+a.c+b.d-a.b-a.c-d.b-d.c
=(a.c-a.c)+(b.d-d.b)+a.d+a.c-a.b-d.c
=0+0+(a-c).(d-b)
=(a-c).(d-b)
1 ) a(b + c) - a(b + d)
= ab + ac - ab - ad
= (ab - ab) + (ac - ad)
= a(c - d) ( đpcm )
2 ) a(b - c) + a(d + c)
= ab - ac + ad + ac
= (ab + ad) - ac + ac
= a(b + d)
Ta có : VT = a ( b + c ) - a ( b + d )
= ab + ac - ab - ad
= ac - ad
= a ( c - d ) = VP
=> a ( b + c ) - a ( b + d ) = a ( c - d ) ( đpcm )
b, Ta có : VT = a ( b - c ) + a ( d + c )
= ab - ac + ad + ac
= ab + ad
= a ( b + d )
=> a ( b - c ) + a ( d + c ) = a ( b + d ) ( đpcm )
a(b+c) - a(b+d)=a(c-d)
VP= a(b+c) - a(b+d)
= ab+ac-ab-ad
= ac-ad
=a(c-d)
Suy ra VP=VT
a(b-c)+a(d+c)=a(b+d)
VP= a(b-c)+a(d+c)
= ab-ac+ad+ac
= ab+ad
= a(b+d)
Suy ra VP=VT
a) \(VT=a\left(b-c\right)-a\left(b+d\right)=a\left(b-c-b-d\right)=-a\left(c+d\right)=VP\)
b) \(VT=\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)=ac+ad+bc+bd-ab-ac-bd-dc\)
\(=ad+bc-ab-dc=a\left(d-b\right)-c\left(d-b\right)=\left(d-b\right)\left(a-c\right)=VP\)
p/s: chúc bạn học tốt
a) \(a\left(b-c\right)-a\left(b+d\right)=ab-ac-ab-ad=-ac-ad=-a\left(c+d\right)\)
=> ĐPCM
b) \(\left(a+b\right)\left(c+d\right)-\left(a+d\right)\left(b+c\right)\)
= a.(c+d)+b(c+d)-[a(b+c)+d(b+c)]
= ac+ad+bc+bd-(ab+ac+bd+cd)
= ac+ad+bc+bd-ab-ac-bd-cd
= ad+bc-ab-cd
= a(d-b)-c(d-b)
= (a-c)(d-b)
=> ĐPCM
a) (a-b+c)-(a+c)
= a-b+c-a-c
= (a-a) + (c-c) - b
= -b
=> a) (a-b+c)-(a+c) = -b
b) a(b+c)-a(b+d)
= ab+ac - ab + ad
= (ab-ab) + ac + ad
= 0 + a(c+d)
= a(c+d)
phần b mk thấy đề sai