Ta có:
$b.x^2 = a.y^2$
$\Leftrightarrow \dfrac{x^2}{a}=\dfrac{y^2}{b}\qquad \quad (a,b\ne 0)$
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
$\dfrac{x^2}{a}=\dfrac{y^2}{b} =\dfrac{x^2 + y^2}{a + b}=\dfrac{1}{a + b}$
+) $\dfrac{x^2}{a}=\dfrac{1}{a + b}$
$\to \left(\dfrac{x^2}{a}\right)^{1010}=\dfrac{1}{(a +b)^{1010}}$
$\to \dfrac{x^{2020}}{a^{1010}}=\dfrac{1}{(a +b)^{1010}}$
+) $\dfrac{y^2}{b}=\dfrac{1}{a + b}$
$\to \left(\dfrac{y^2}{b}\right)^{1010}=\dfrac{1}{(a +b)^{1010}}$
$\to \dfrac{y^{2020}}{b^{1010}}=\dfrac{1}{(a +b)^{1010}}$
Do đó:
$\dfrac{x^{2020}}{a^{1010}} + \dfrac{y^{2020}}{b^{1010}} = \dfrac{1}{(a +b)^{1010}} + \dfrac{1}{(a +b)^{1010}} = \dfrac{2}{(a +b)^{1010}}$
