Question

Задание на фото 100б

Answer 1

$\displaystyle\bf\\8\sqrt{12+16x-16x^{2} } +x(4-4x)=33\\\\8\sqrt{4\cdot(3+4x-4x^{2} )} +(4x-4x^{2} )=33\\\\8\cdot 2\cdot\sqrt{3+4x-4x^{2} } +(4x-4x^{2} )=33\\\\16\cdot\sqrt{3+4x-4x^{2} } +(4x-4x^{2} )=33\\\\\sqrt{3+4x-4x^{2} }=m \ , \ m\geq 0\\\\\Big(\sqrt{3+4x-4x^{2} }\Big)^{2} =m^{2} \\\\3+4x-4x^{2} =m^{2} \\\\4x-4x^{2} =m^{2}-3\\\\16m+m^{2} -3=33\\\\m^{2} +16m-36=0\\\\D=16^{2} -4\cdot(-36)=256+144=400=20^{2} \\\\\\m_{1} =\frac{-16+20}{2} =\frac{4}{2} =2$

$\displaystyle\bf\\m_{2} =\frac{-16-20}{2} =-18 < 0-neyd\\\\\sqrt{3+4x-4x^{2} } =2\\\\3+4x-4x^{2} =4\\\\4x^{2} -4x+1=0\\\\(2x-1)^{2} =0\\\\2x-1=0\\\\x_{0} =0,5\\\\36x_{0} =36\cdot 0,5=18\\\\Otvet \ : \ 18$