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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:-a(c-d)-d(a+c)
=-ac+ad-da-dc
=-ac-dc
=-c(a+d) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-a.\left(c-d\right)-d.\left(a+c\right)=-ac+ad-ad-dc=\left(ad-ad\right)-ac-dc=-ac-dc=-\left(ac+dc\right)=-c.\left(a+d\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(a.\left(b+c\right)-b.\left(a-c\right)=a.b+a.c-b.a+b.c=a.c+b.c=c.\left(a+b\right)\)
b) \(a.\left(b-c\right)-a.\left(b+d\right)=a.b-a.c-a.b-a.d=-a.c-a.d=-a.\left(c+d\right)\)
ĐPCM
a)Xét VT(vế trái)=a.(b+c)-b.(a-c) b)Xét VT=a(b-c)-a(b+d)
- =ab+ac-ba+bc. =ab-ac-ab-ad
- =c.(a+b)=VP(vế phải). =-ac-ad
- =-a(c+d)=VP
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(-\left(-m+n+p\right)+\left(n+p-3\right)=\left(n-p+8\right)-\left(11-m+n\right)+p\\ \Leftrightarrow m-n-p+n+p-3=n-p+8-11+m-n+p\\ \Leftrightarrow\left(n-n\right)+\left(p-p\right)+m-3=\left(n-n\right)+\left(p-p\right)+m+\left(8-11\right)\\ \Leftrightarrow m-3=m+\left(-3\right)\\ \Leftrightarrow m-3=m-3\\ \Leftrightarrow0=0\left(\text{luôn đúng}\right)\)
Ta được đpcm
Bài 2:
\(A-B=\left(b-c-4\right)-\left(b-a\right)\\ A-B=b-c-4-b+a\\ A-B=\left(b-b\right)+a-c-4\\ A-B=a-c-4\left(1\right)\)
\(C+D=\left(-b-c+1\right)+\left(a+b-5\right)\\ C+D=-b-c+1+a+b-5\\ C+D=\left(b-b\right)+a-c+\left(1-5\right)\\ C+D=a-c+\left(-4\right)\\ C+D=a-c-4\left(2\right)\)
(1) (2) \(\Rightarrow A-B=C+D\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Vế trái = a(b + c) - b(a - c)
= ab + ac - ba + bc
= ac + bc = (a + b)c = vế phải
![](https://rs.olm.vn/images/avt/0.png?1311)
-(-a+b+c)+(b+c-1)=(b-c+6)-(7-a-b)+c
a-b-c+b+c-1=b-c+6-7+b+c
a-1=2c-1
a=2c
Ta có :
- ( -a + b + c ) + ( b + c -1 )
= a - b - c + b + c - 1
= ( b - b ) + ( c - c ) + a - 1
= 0 + 0 + a -1
= a - 1
Ta có :
( b - c + 6 ) - ( 7 - a + b ) + c
= b - c + 6 - 7 + a - b + c
= 0 + 0 + a - 1
= a - 1
Vì a - 1 = a - 1
=> - ( a + b + c ) + ( b + c - 1 ) = ( b - c + 6 ) - ( 7 - a + b ) + c ( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
Vế trái = (a + b)(a - b)
= a.a + b.a - a.b - b.b
= a2 - b2 = vế phải
=-ac+ad-ad-dc
=(ad-ad)-ac-dc
=0-(ac+cd)
=-(ac+cd)
=-c(a+d)