Question

пусть (x;y) - решение системы, найдите х+2у

$\left \{ {{\sqrt{x} -\sqrt{y} = \frac{1}{2} \sqrt{xy}} \atop {x+y=5}} \right.$

Answer 1

$\left \{ {{\sqrt{x}-\sqrt{y}=\frac{1}{2}\sqrt{xy}} \atop {x+y=5}} \right.\; \; t=\sqrt{x}\geq 0\; ,\; \; p=\sqrt{y}\geq 0\; ,\; \; x\geq 0\; ,\; y\geq 0\\\\\left \{ {{t-p=\frac{1}{2}tp} \atop {t^2+p^2=5}} \right.\; \; \left \{ {{tp=2(t-p)\qquad \; \; } \atop {t^2+p^2-2tp+2tp=5}} \right.\; \; \left \{ {{tp=2(t-p)} \atop {(t-p)^2+2tp=5}} \right. \left \{ {{tp=2(t-p)\qquad \quad } \atop {(t-p)^2+4(t-p)-5=0}} \right.\\\\a)\; \; (t-p)^2+4(t-p)-5=0\; \; ,\\\\(t-p)_1=-5\; \; ili\; \; \; (t-p)_2=1\; \; (teorema\; Vieta)$

$b)\; \; \left \{ {{t-p=-5} \atop {t-p=\frac{1}{2}\, tp}} \right.\; \; \left \{ {{t=p-5\qquad } \atop {-5=\frac{1}{2}(p^2-5p)}} \right.\; \; \left \{ {{t=p-5\qquad } \atop {p^2-5p+10=0}} \right. \\\\p^2-5p+10=0\; ,\; \; D=25-40<0\; \; \to \; \; p\in \varnothing \\\\c)\; \; \left \{ {{t-p=5} \atop {t-p=\frac{1}{2}\, tp}} \right.\; \; \left \{ {{t=p+5} \atop {p^2+5p-10=0}} \right. \\\\p^2+5p-10=0\; ,\; D=65\; \; ,\; \; p_{1,2}=\frac{-5\pm \sqrt{65}}{2}\\\\p\geq 0\; \; \to \; \; p=\frac{-5+\sqrt{65}}{2}\; \; ,\; \; t=\frac{5+\sqrt{65}}{2}$

$d)\; \; \sqrt{y}=\frac{-5+\sqrt{65}}{2}\; \; ,\; \; y=(\frac{-5+\sqrt{65}}{2})^2=\frac{90-10\sqrt{65}}{4}=\frac{45-5\sqrt{65}}{2}\\\\\sqrt{x}=\frac{5+\sqrt{65}}{2}\; \; ,\; \; x=(\frac{5+\sqrt{65}}{2})^2=\frac{45+5\sqrt{45}}{2}\\\\Otvet:\; \; (\frac{45+5\sqrt{45}}{2}\, ;\, \frac{45-5\sqrt{45}}{2})\; .$