Question

Решите уравнение (√2(x²-6x)+19)+(√3x²-18x+36)=-2x²+12x-14

Answer 1

$sqrt{2(x^2-6x)+19}+sqrt{3(x^2-6x)+36} = -2(x^2-6x)-14\ 3x^2-18x+36=3(x^2-6x+12)\ x^2-6x+12=t\ sqrt{2t-5}+sqrt{3t}=-2t+10\ 2t-5+3t+2sqrt{6t^2-15t}=10-2t\ sqrt{6t^2-15t}*2=15-7t\ (6t^2-15t)*4=225-210t+49t^2 24t^2-60t=225-210t+49t^2\ 25t^2-150t+225=0\ t^2-6t+9=0 \ (t-3)^2=0\ t=3\ x^2-6x+12=3\ x^2-6x+9=0\ (x-3)^2=0\ x=3$