7. Solve the equation $\sqrt{x}\left(x^{2}-3 x-4\right)^{3}+\sqrt{4-x}\left(x^{2}-8 x\right)^{5}=0$.
Answer: $x=0 ; x=4$.

Solution The domain of definition (ODZ) of the given equation is the interval [0;4]. On this interval, the quadratic trinomials $x^{2}-3 x-4 n x^{2}-8 x$ do not exceed 0. The sum of two non-positive terms equals 0 only if each of them is 0. The first one is 0 at $x=-1,0,4$. The second one is 0 at $x=0,4,8$. Only 0 and 4 coincide.

## Variant 1-a