Question

Решите уравнение, пожалуйста

Answer 1

$\displaystyle\\ODZ:\\\\\sqrt{4x-3x^2-1}\geq 0\\\\\sqrt{4x-3x^2-1}\geq 0\\\\4x-3x^2-1\geq 0\\\\3x^2-4x+1\leq 0\\\\D=16-4*3*1=16-12=4\\\\\sqrt{D}=\sqrt{4}=2\\\\x_1=\frac{4+2}{6}=\frac{6}{6}=1\\\\x_2=\frac{4-2}{6}=\frac{2}{6}=\frac{1}{3}\\\\(x-1)(x-\frac{1}{3})\leq 0\\\\+++++[1]-----[\frac{1}{3}]+++++\\\\x\in\bigg[\frac{1}{3};1\bigg]$

$\displaystyle\\(7x+2)\sqrt{4x-3x^2-1}=0\\\\\left[ \begin{gathered} 7x+2=0;\ \ \ \ \ \ (1)\\ \sqrt{4x-3x^2-1}=0\ (2), \\ \end{gathered} \right.\\\\\\(1):\\\\7x+2=0\\\\7x=-2\\\\x=-\frac{2}{7}\\\\\\(2):\\\\\sqrt{4x-3x^2-1}=0\\\\4x-3x^2-1=0\\\\3x^2-4x+1=0\\\\D=16-4*3*1=16-12=4\\\\\sqrt{D}=\sqrt{4}=2\\\\x_1=\frac{4+2}{6}=\frac{6}{6}=1\\\\x_2=\frac{4-2}{6}=\frac{2}{6}=\frac{1}{3}\\\\\\x=-\frac{2}{7}\notin ODZ\\\\\boxed{x=1\ \ \ x=\frac{1}{3} }$