Cho hai số thực a , b

- Hướng dẫn giải

Với ab = 1 , a + b ¹ 0, ta có:

$\begin{array}{l}P=\frac{a^3+b^3}{{(a+b)}^3{\left(ab\right)}^3}+\frac{3(a^2+b^2)}{{(a+b)}^4{\left(ab\right)}^2}+\frac{6(a+b)}{{(a+b)}^5\left(ab\right)}\\ =\frac{a^3+b^3}{{(a+b)}^3}+\frac{3(a^2+b^2)}{{(a+b)}^4}+\frac{6(a+b)}{{(a+b)}^5}\\ =\frac{a^2+b^2-1}{{(a+b)}^2}+\frac{3(a^2+b^2)}{{(a+b)}^4}+\frac{6}{{(a+b)}^4}\\ =\frac{(a^2+b^2-1){(a+b)}^2+3(a^2+b^2)+6}{{(a+b)}^4}\\ =\frac{(a^2+b^2-1)(a^2+b^2+2)+3(a^2+b^2)+6}{{(a+b)}^4}\\ =\frac{{(a^2+b^2)}^2+4(a^2+b^2)+4}{{(a+b)}^4}\\ =\frac{{(a^2+b^2+2)}^2}{{(a+b)}^4}\\ =\frac{{(a^2+b^2+2ab)}^2}{{(a+b)}^4}\\ =\frac{{\left[{(a+b)}^2\right]}^2}{{(a+b)}^4}\\ =1\end{array}$

Vậy P = 1, với ab = 1 , a+b ¹ 0.