Question

Найти действительные корни уравнения (2x+1)(3x+2)(6x+1)(x+1)=210

Answer 1

$(2x+1)(3x+2)(6x+1)(x+1)=210\ (2x+1)(3x+2) = 6x^x+7x+2\ (6x+1)(x+1) = 6x^2+7x+1\ (6x^x+7x+2)(6x^2+7x+1)=210\ 6x^2+7x=t\ (t+2)(t+1)=210\ t^2+3t-208=0\ D=9+208*4=841\ t_{1,2}=frac{-3pm29}{2} = -16;13\ 6x^2-7x+16=0\ D=49-6*16*4<0\ 6x^2-7x-13=0\ D=49+13*6*4=161\ x_{1,2}=frac{-7pm19}{12}= -frac{13}{6};1$