Question

Помогите решить с проверкой ОДЗ!

Answer 1

Ответ:

Пошаговое объяснение:

$\sqrt{x+8}=9\\(\sqrt{x+8}) ^{2} = 9^{2} \\x+8=81\\x=81-8\\x=73\\\\\sqrt{73+8} = 9\\\sqrt{81} =9\\9=9\\\\\\\sqrt{3-2x}=\sqrt{x+4} \\(\sqrt{3-2x})^{2} =(\sqrt{x+4})^{2} \\3-2x=x+4\\-2x-x=4-3\\-3x=1\\x=\frac{1}{-3} \\\\\\\sqrt{3x+1} =x-1\\(\sqrt{3x+1})^{2} =(x-1)^{2} \\3x+1=x^{2} -2x+1\\x^{2} -2x+1-3x-1=0\\x^{2} -5x=0\\x(x-5)=0\\x_{1}=0\\x_{2}= 5\\\\\sqrt{3*5+1} =5-1\\\sqrt{16} =4\\4=4\\Otvet: 5$