a) \(\frac{a^2m-a^2n-b^2n+b^2m}{a^2+b^2}=\frac{a^2\left(m-n\right)+b^2\left(m-n\right)}{a^2+b^2}\)

\(=\frac{\left(m-n\right)\left(a^2+b^2\right)}{a^2+b^2}=m-n\)

b) \(\frac{\left(ab+bc+cd+ad\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-b\right)}\)

\(=\frac{\left[b.\left(a+c\right)+d.\left(a+c\right)\right].abcd}{ac+bc+da+db+ab-b^2-ca+bc}\)

\(=\frac{\left(a+c\right)\left(d+b\right)abcd}{2bc+da+db+ab-b^2}\)

a ) \(A=\frac{ax^2\left(a-x\right)-a^2x\left(x-a\right)}{3a^2-3x^2}=\frac{ax\left(a-x\right)\left(a+x\right)}{3\left(a-x\right)\left(a+x\right)}=\frac{ax}{3}\)

Thay \(a=\frac{1}{2};x=-3\), ta có :

\(A=\frac{\frac{1}{2}.-3}{3}=-\frac{1}{2}\)

b ) \(B=\frac{\left(ab+bc+cd+da\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}=\frac{\left[\left(ab+ad\right)+\left(bc+cd\right)\right]abcd}{ca+cb+da+db+ba-bd-ca+cd}\)

\(=\frac{\left[a\left(b+d\right)+c\left(b+d\right)\right]abcd}{ba+da+cb+cd}=\frac{\left(b+d\right)\left(a+c\right)abcd}{\left(b+d\right)\left(a+c\right)}=abcd\)

Thay \(a=-3;b=-4;c=2;d=3\), ta có :

\(B=\left(-3\right).\left(-4\right).2.3=72\)

\(A=\dfrac{-\left(ac+bc+ad+bd\right)-\left(cd-ca-bd+ba\right)}{\left(ab+bc+cd+ad\right)\cdot abcd}\)

\(=\dfrac{-ac-bc-ad-bd-cd+ca+bd-ba}{\left(ab+bc+cd+ad\right)\cdot abcd}\)

\(=\dfrac{-bc-ad-cd-ba}{\left(ab+bc+cd+ad\right)\cdot abcd}=-\dfrac{1}{abcd}\)

Bất đẳng thức trên đúng với mọi số thực a, b, c. Ai có thể chứng minh?