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1) \(=ac+ad+bc+bd-ab-ac-db-dc=ad+bc-dc-ab=d\left(a-c\right)-b\left(a-c\right)=\left(a-c\right)\left(d-b\right)\)
2) \(=ac-ad+bc-bd-ac-ad+bc+bd=2bc-2ad=2\left(bc-ad\right)\)
3) \(\left(a+b\right)\left(a+b\right)-\left(a-b\right)\left(a-b\right)=a^2+2ab+b^2-a^2+2ab-b^2=4ab\)
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)
a)-(-a+c-d)-(c-a+d)=a-c+d-c+a-d=(a+a)-(c+c)+(d-d)=2a-2c=2(a-c)
b)-(a+b-c+d)+(a-b-c-d)=-a-b+c-d+a-b-c-d=(-a+a)-(b+b)+(c-c)-(d+d)=0-2b+0-2d=-2(b-d)
c)a(b-c-d)-a(b+c-d)=ab-ac-ad-ab-ac+ad=(ab-ac)-(ac+ac)-(ad-ad)=2ac
d)đề sai
e)(a+b)(c-d)-(a-b)(c+d)=ac+b-ad+b-(ac-b+ad-b)=ac+b-ad+b-ac+b-ad+b=(ac-ac)+(b+b+b+b)-(ad+ad)=4b-2ad=2(2b-ad)
f)(a+b)2-(a-b)2=a2+2ab+b2-(a2-2ab+b2)=a2+2ab+b2-a2+2ab-b2=(a2-a2)+(2ab+2ab)+(b2-b2)=4ab
mk k chắc đâu
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
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
- a-c+d-c+d-d=a-2c+d
- -a-b+c-d+a -b-c-d=-2b-2d
- ab-ac-ad-ab-ac+ad=-2ab-2ac
- ac+ad+bc+bd-ab-ac-bd-cd=ad+bc+bd-ab-bd-cd