Question

Chứng minh đẳng thức sau :

( a+b)(b+c)-(c+d)(d+a)-(a+c)(b-d) =b² - d²

Answer 1

=(ab+ac+b^2+bc)- (cd+ca+d^2+ad)-(ab-ad+cb-cd)

=ab+ac+b^2+bc-cd-ca-d^2-ad-ab+ad-cb+cd

=b^2-d^2

Answer 2

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

\(=\left(ab+ac+b^2+bc\right)-\left(cd+ac+d^2+ad\right)-\left(ab-ad+bc-cd\right)\)

\(=ab+ac+b^2+bc-cd-ac-d^2-ad-ab+ad-bc+cd\)

\(=b^2-d^2.\)

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