$\frac{sin^4x}{a}+\frac{cos^4x}{b}=\frac{1}{a+b}$$\Leftrightarrow sin^4x.\frac{a+b}{a}+cos^4x.\frac{a+b}{b}=1$
$\Leftrightarrow sin^4x+sin^4x.\frac{b}{a}+cos^4x+cos^4x.\frac{a}{b}=(sin^2x+cos^2x)^2$
$\Leftrightarrow sin^4x.\frac{b}{a}+cos^4x.\frac{a}{b}=2sin^2x.cos^2x$
$\Leftrightarrow b^2sin^4x+a^2cos^4x=2ab.sin^2x.cos^2x$
$\Leftrightarrow (bsin^2x-acos^2x)^2=0$
$\Leftrightarrow bsin^2x=acos^2x$
$\Leftrightarrow \frac{sin^2x}{a}=\frac{cos^2x}{b}=\frac{sin^2x+cos^2x}{a+b}=\frac{1}{a+b}$
$\Leftrightarrow \begin{cases}sin^2x=\frac{a}{a+b} \\ cos^2x=\frac{b}{a+b} \end{cases}$
Thay vào biểu thức đã cho
$\frac{sin^{10}x}{a^4}+\frac{cos^{10}x}{b^4}$
$=\frac{1}{a^4}.\frac{a^5}{(a+b)^5}+\frac{1}{b^4}.\frac{b^5}{(a+b)^5}$
$=\frac{a}{(a+b)^5}+\frac{b}{(a+b)^5}=\frac{1}{(a+b)^4}$