Question

Решите ,пожалуйста.Даю 23 балла

Answer 1

$x^{2} +2\sqrt{x^{2}-3x+11 }=3x+4\\\\x^{2}-3x-4+2\sqrt{x^{2}-3x+11 } =0\\\\(x^{2}-3x+11) -15+2\sqrt{x^{2}-3x+11 }=0\\\\\sqrt{x^{2}-3x+11 }=m,m\geq 0 \Rightarrow x^{2}-3x+11=m^{2} \\\\m^{2}-15+2m=0\\\\m^{2}+2m-15=0\\\\m_{1}=3,teorema.Vieta\\\\m_{2}=-5-neyd\\\\\sqrt{x^{2}-3x+11 }=3\\\\(\sqrt{x^{2}-3x+11} )^{2}=3^{2}\\\\x^{2}-3x+11=9\\\\x^{2}-3x+2=0\\\\x_{1}=1\\\\x_{2}=2\\\\Proverka:\\\\1)\sqrt{1^{2}-3*1+11 }=\sqrt{9}=3>0-verno\\\\2)\sqrt{2^{2}-3*2+11 }=\sqrt{9}=3>0-verno$

$Otvet:\boxed{1;2}$

Answer 2

$x^2+2\sqrt{x^2-3x+11}=3x+4\\x^2-3x+1\geq 0\Rightarrow \forall x\in \mathbb{R}\\x^2-3x-4+2\sqrt{x^2-3x+11}=0\\x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\\\sqrt{x^2-3x+11}=t,t\geq 0\Rightarrow \\\Rightarrow t^2+2t-15=0\\D_1=1+15=16\\t=-1\pm 4\Rightarrow t=\begin{Bmatrix}-5;3\end{Bmatrix}\Rightarrow \\\Rightarrow \sqrt{x^2-3x+11}=3\Leftrightarrow \\\Leftrightarrow x^2-3x+11=9\Leftrightarrow \\\Leftrightarrow x^2-3x+2=0\\x=\begin{Bmatrix}1;2\end{Bmatrix}$