(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

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

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

\(=ad+bc-ab-dc\)

\(=\left(ad-ab\right)+\left(bc-dc\right)\)

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

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

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

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

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

\(=-2c\)

giải hẳn ra giùm mik vs

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

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

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

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

\(D=4b-3a\)

\(\text{- ( a - b - c ) + ( b - c + d) - ( -a + b + d) }\)

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

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

\(=0+b+0+0\)

\(=b\)

a) a ≠ 0 ,    a ≠   − 5  

b) Ta có A = a 3 + 4 a 2 − 5 a 2 a ( a + 5 ) = a ( a − 1 ) ( a + 5 ) 2 a ( a + 5 ) = a − 1 2  

c) Thay a = -1 (TMĐK) vào a ta được A = -1

d) Ta có A = 0 Û a = 1 (TMĐK)

D

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