Question

помогите пожалуйста разобраться​

Answer 1

$\displaystyle\bf\\\sqrt{x-5} +\sqrt{10-x} =3\\\\ODZ:\\\left\{\begin{array}{ccc}x-5\geq 0\\10-x\geq 0\end{array}\right\\\\\\\left\{\begin{array}{ccc}x\geq 5\\x\leq 10\end{array}\right \ \ \Rightarrow \ x\in\Big[5 \ ; \ 10\Big]\\\\\\\Big(\sqrt{x-5}\Big)^{2} =\Big(3-\sqrt{10-x} \Big)^{2} \\\\x-5=9-6\sqrt{10-x} +10-x\\\\6\sqrt{10-x} =24-2x\\\\\Big(3\sqrt{10-x} \Big)^{2} =\Big(12-x\Big)^{2} \\\\9(10-x)=144-24x+x^{2}\\\\90-9x=144-24x+x^{2} \\\\x^{2} -15x+54=0\\\\Teorema \ Vieta:$

$\displaystyle\bf\\x_{1} =6 \ \ ; \ \ x_{2} =9$