\(A=\left(a-b-c-d\right)+\left(b-c+d-a\right)\)

\(=a-b-c-d+b-c+d-a\)

\(=-2c\)

giải hẳn ra giùm mik vs

(a+b).(c+d)-(a+d).(b+c)

=(-c-b).d+b.c+b^2

(a+b).(c+d)-(a+d).(b+c)

=ac+ad+bc+bd-ab-ac-bd-dc

=(ac-ac)+(bd-bd)+(ad-ab)+(bc-dc)

=0+0+a.(d-b)+c.(b-d)

=a.(d-b)+c.(b-d)

Hok tốt

\(-d-c+2b+2a\)

a) - ( - a + c – d ) – ( c – a + d ) 

= a - c - d - c + a + d

= (a + a) + (-c - c) + (-d + d)

= 2a - 2c               

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)

= -2b - 2d

a) - ( - a + c - d) - ( c - a + d )

= a - c + d - c + a - d 

= 2a

b) - ( a+ b - c + d ) + ( a -b -c -d )

= - a-b+c-d+a-b-c-d

=-2d -2b

c) a(b-c-d) - a(b+c-d)

= a(b-c-d-b-c+d)

= ab-ac-ad-ab-ac+ad

= -2ab-2ac

d) (a+b)(c+d)-(a+d)(b+c)

= ac+ad+bc+bd - (ab+ac+bd+cd)

= ac+ad+bc+bd-ab-ac-bd-cd

=ad+bc-ab-cd

\(D=\left(a+b-c\right)-\left(a-b+c\right)+\left(b+c-a\right)-\left(a-b-c\right)\)

\(D=a+b-c-a+b-c+b+c-a-a+b+c\)

\(D=\left(a-a-a-a\right)+\left(b+b+b+b\right)+\left(c+c-c-c\right)\)

\(D=4b-3a\)

\(A=\left[\left(a+b\right)+\left(c+d\right)\right]^2+\left[\left(a+b\right)-\left(c+d\right)\right]^2+\left[\left(a-b\right)+\left(c-d\right)\right]^2+\left[\left(a-b\right)-\left(c-d\right)\right]^2\)

Ta có

\(\left[\left(a+b\right)+\left(c+d\right)\right]^2=\left(a+b\right)^2+2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)

\(\left[\left(a+b\right)-\left(c+d\right)\right]^2=\left(a+b\right)^2-2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)

\(\left[\left(a-b\right)+\left(c-d\right)\right]^2=\left(a-b\right)^2+2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)

\(\left[\left(a-b\right)-\left(c-d\right)\right]^2=\left(a-b\right)^2-2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)

\(A=2\left(a+b\right)^2+2\left(a-b\right)^2+2\left(c+d\right)^2+2\left(c-d\right)^2\)

\(A=2\left(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2\right)\)

\(A=4\left(a^2+b^2+c^2+d^2\right)\)