Question

если знаете не проходите мимо

Answer 1

$(\sqrt[4]{x^{2} +3x}) ^{4} =(\sqrt[4]{21-x})^{4} \\x^{2} +3x=21-x\\x^{2} +3x-21+x=0\\x^{2} +4x-21=0\\D=4^{2}-4*1*(-21)=16+84=100\\x_{1,2}=\frac{-4б\sqrt{100} }{2*1} =\frac{-4б10}{2}\\x_{1}=\frac{-4+10}{2}=\frac{6}{2}=3\\x_{2} =\frac{-4-10}{2}=\frac{-14}{2} =-7x^{2} +3x\geq 0x^{2} _{1} +3x_{1}\geq 0\\3^{2} +3*3\geq 0\\9+9\geq 0\\18\geq 0x_{2} ^{2} +3x_{2} \geq 0\\(-7)^{2}+3*(-7)\geq 0\\49-21\geq 0\\28\geq 0x_{1} +x_{2} =3+(-7)=3-7=-4$

ответ. -4