Question

Cho ab=1. Chứng minh rằng:

\(a^5+b^5=\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\)

Answer 1

Biến đổi VP ta có :

\(VP=\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\)

\(=a^5+a^3b^2+a^2b^3+b^5-\left(a+b\right)\)

\(=a^5+a.\left(ab\right)^2+b.\left(ab\right)^2+b^5-\left(a+b\right)\)

\(=a^5+a+b+b^5-\left(a+b\right)\) (vì \(ab=1\))

\(=a^5+b^5=VT\)(đpcm)

Answer 2

Biến đổi vế phải :
\(\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)=a^5+b^5+a^3b^2+a^2b^3-\left(a+b\right) \)

\(=a^5+b^5+a^2b^2\left(a+b\right)-\left(a+b\right)\)

\(=a^5+b^5+\left(a+b\right)-\left(a+b\right)\)(vì ab=1)

\(=a^5+b^5\)