Алгебра.............

$\sqrt{x-5}+ \sqrt{10-x} =3 \ (x\in[5;10]) \\\ ( \sqrt{x-5}+ \sqrt{10-x})^2 =3^2 \\\ x-5+10-x+2 \sqrt{(x-5)(10-x)} =9 \\\ 5+2 \sqrt{(x-5)(10-x)} =9 \\\ 2 \sqrt{(x-5)(10-x)} =4 \\\ \sqrt{(x-5)(10-x)} =2 \\\ ( \sqrt{(x-5)(10-x)} )^2=2^2 \\\ (x-5)(10-x)=4 \\\ 10x-x^2-50+5x-4=0 \\\ x^2-15x+54=0 \\\ x^2-6x-9x+54=0 \\\ x(x-6)-9(x-6)=0 \\\ (x-6)(x-9)=0 \\\ x_1=6 ; \ x_2=9$
Ответ: 6 и 9

$\sqrt{3x+1} -2- \sqrt{x+1} =0 \ (x \geq - \frac{1}{3} ) \\\ \sqrt{3x+1} =2+ \sqrt{x+1} \\\ ( \sqrt{3x+1})^2 =(2+ \sqrt{x+1})^2 \\\ 3x+1=4+x+1+4 \sqrt{x+1} \\\ 2x-4=4 \sqrt{x+1} \\\ x-2=2 \sqrt{x+1} \ (x \geq 2) \\\ (x-2)^2=(2 \sqrt{x+1})^2 \\\ x^2-4x+4=4(x+1) \\\ x^2-4x+4-4x-4=0 \\\ x^2-8x=0 \\\ x(x-8)=0 \\\ x_1 \neq 0<2 ; \ x_2=8$
Ответ: 8