Question

Cho các số thực a, b thỏa mãn a + b = 1 và ab khác 0. Tính

\(P=\frac{a}{b^3-1}-\frac{b}{a^3-1}+\frac{2\left(a-b\right)}{a^2b^2+3}\)

giúp mk với mau nha

Answer 1

\(P=\frac{1-b}{\left(b-1\right)\left(b^2+b+1\right)}-\frac{1-a}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{2\left(a-b\right)}{a^2b^2+3}\)

\(P=\frac{1}{a^2+a+1}-\frac{1}{b^2+b+1}+\frac{2\left(a-b\right)}{a^2b^2+3}\)

\(P=\frac{b^2+b+1-a^2-a-1}{a^2b^2+a^2b+a^2+ab^2+ab+a+b^2+b+1}+\frac{2\left(a-b\right)}{a^2b^2+3}\)

\(P=\frac{\left(b-a\right)\left(b+a\right)+b-a}{a^2b^2+ab\left(a+b\right)+ab+a^2+b^2+2}+\frac{2\left(a-b\right)}{a^2b^2+3}\)

\(P=\frac{2\left(b-a\right)}{a^2b^2+a^2+2ab+b^2+2}+\frac{2\left(a-b\right)}{a^2b^2+3}\)

\(P=\frac{-2\left(a-b\right)}{a^2b^2+\left(a+b\right)^2+2}+\frac{2\left(a-b\right)}{a^2b^2+3}\)

\(P=\frac{-2\left(a-b\right)}{a^2b^2+3}+\frac{2\left(a-b\right)}{a^2b^2+3}=0\)