Question

Помогите решить пожалуйстa

Answer 1

$tt (x^2+4)^2+5(x^2+4)-24=0\x^2+4=y\y^2+5y-24=0\D=b^2-4ac=25+4cdot 24=121=11^2\y_{1,2}=frac{-bpmsqrt{D}}{2a}=frac{-5pm11}{2}=left |{ {{3} atop {-8}} right. \left [{ {{x^2+4=3} atop {x^2+4=-8}} right. left [{ {{x^2neq -1} atop {x^2neq -12}} right.$

Ответ: корней нет

$tt (x^2+x+2)(x^2+x-5)=-10\x^2+x+2=z\z(z-7)+10=0\z^2-7z+10=0\D=49-40=9=3^2\z_{1,2}=frac{7pm 3}{2}=left |{ {{5} atop {2}} right. \left [{ {{x^2+x+2=5} atop {x^2+x+2=2}} right.$

$tt x^2+x+2=5\x^2+x-3=0\D=1+12=13=(sqrt{13})^2\x_{1,2}=frac{-1pmsqrt{13}}{2}$

$tt x^2+x+2=2\x^2+x=0\x(x+1)=0\left [{ {{x=-1} atop {x=0}} right.$

Ответ: $tt frac{-1-sqrt{13}}{2};frac{-1+sqrt{13}}{2};-1;0$