Question

(x^2-64)^2+(x^2+5x-24)^2=0

Answer 1

$(x^{2} -64)^{2} +(x^{2} +5x-24)^{2}=0$

$a^{2} +b^{2} =0$, если a=0 и b=0

$1) x^{2} -64=0\\x^{2} =64\\x_{1} =8; x_{2} =-8$

$2) x^{2} +5x-24=0\\D=25+24*4=121\\x_{1}= \frac{-5+11}{2}=3\\x_{2} =\frac{-5-11}{2} =-8$

Ответ: -8; 3; 8