1. Solve the equation

$$
\sqrt{x-9}+\sqrt{x+12}-\sqrt{x-3}-\sqrt{x-4}=0
$$
Solution. Squaring the equation $\sqrt{x-9}+\sqrt{x+12}=\sqrt{x-3}+\sqrt{x-4}$ yields

$$
x-9+2 \sqrt{x^{2}+3 x-108}+x+12=x-3+2 \sqrt{x^{2}-7 x+12}+x-4
$$

i.e.

$$
\sqrt{x^{2}+3 x-108}=\sqrt{x^{2}-7 x+12}-5
$$

Squaring again, we get

$$
x^{2}+3 x-108=x^{2}-7 x+12-10 \sqrt{x^{2}-7 x+12}+25
$$

i.e.

$$
2 \sqrt{x^{2}-7 x+12}=29-2 x
$$

Squaring once more, we get

$$
4 x^{2}-28 x+48=841-116 x+4 x^{2} \text { i.e. } 88 x=793
$$

Since $\frac{793}{88}-9=\frac{1}{88}>0$, the desired solution is $x=\frac{793}{88}$.