Question

алгебра.............

Answer 1

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