d*)(a+b)(c+d)-(a+d)(b+c)
e*)(a+b)(c-d)-(a-b)(c+d)
f*)(a+b)2-(a-b)2
K
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NT
0
12 tháng 1 2021
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)
BT
0
27 tháng 6 2017
\(-(-a+c-d)-(c-a+d)\)
\(=a-c+d-c+a-d\)
\(=(a+a)-(c+c)+(d-d)\)
\(=2a-2c\)
\(-(a+b-c+d)+(a-b-c-d)\)
\(=-a-b+c-d+a-b-c-d\)
\(=(-a+a)-(b+b)+(c-c)-(d+d)\)
\(=0-0+2c-2d\)
\(=2c-2d\)
\(a(b-c-d)-a(b+c-d)\)
\(=a(b-c-d-b+c+d)\)
\(=a.0=0\)
\((a+b)(c+d)-(a+d)(b+c)\)
\(=a(c+d)+b(c+d)-a(b+c)+d(b+c)\)
\(=ac+ad+bc+bd-ab-ac-bd-dc\)
\(=(ac-ac)+(bd-bd)+ad-ab-bc-dc\)
\(=a(d-b)-c(b+d)\)
\((a+b)(c-d)-(a-b)(c+d)\)
\(=a(c-d)+b(c-d)-a(c+d)+b(c-d)\)
\(=ac-ad+bc-bd-ac-ad+bc-bd\)
\(=(ac-ac)-(ad+ad)+(bc+bc)-(bd+bd)\)
\(=2ad+2bc-2bd\)
\((a+b)^2-(a-b)^2\)
\(=a^2+2ab+b^2-a^2+2ab-b^2\)
\(=2ab+2ab=4ab\)
20 tháng 7 2015
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 tháng 11 2016
- 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
\((a+b)(c+d)-(a+d)(b+c)\)
\(=a(c+d)+b(c+d)-a(b+c)-d(b+c)\)
\(=ac+ad+bc+bd-ab-ac-bd-cd\)
\(=(ac-ac)+(bd-bd)+ad+bc-ab-cd\)
\(=c(b-d)+d(a-c)\)
e) tương tự d
\((a+b)^2-(a-b)^2\)
\(=(a^2+2ab+b^2)-(a^2-2ab+b^2)\)
\(=a^2+2ab+b^2-a^2+2ab-b^2\)
\(=4ab\)