Question

Решите систему уравнений

Answer 1

$\begin{cases}\sqrt{x}=a,a\geq 0\\ \sqrt{y}=b,b\geq 0\end{cases}\Rightarrow \begin{cases}a^3+12b^2a=28\\ 8b^3+6a^2b=36\end{cases}\Rightarrow 9\left ( a^3+12ab^2-28 \right )-14\left ( 3a^2b+4b^3-18 \right )=0 \\9a^3+108ab^2-42a^2b-56b^3=0\overset{b\neq 0}{\Leftrightarrow }9\left ( \cfrac{a}{b} \right )^3+108\cdot \cfrac{a}{b}-42\left ( \cfrac{a}{b} \right )^2-56=0\\\cfrac{a}{b}=t\Rightarrow 9t^3-42t^2+108t-56=0\Leftrightarrow 9t^3-6t^2-35t^2+24t+84t-56=0$$9t^3-6t^2-36t^2+24t+84t-56=0\Leftrightarrow 3t^2(3t-2)-12t(3y-2)+28(3t-2)=0\\(3t-2)\underbrace{\left ( 3t^2-12t+28 \right )}_{t\in \mathbb{C}}=0\Rightarrow t=\cfrac{2}{3}\Rightarrow a=\cfrac{2}{3}b\Rightarrow \cfrac{8}{27}b^3+8b^3-28=0\\\cfrac{224}{27}b^3-28=0\Rightarrow b=\cfrac{3}{2}\Rightarrow a=\cfrac{2}{3}\cdot \cfrac{3}{2}=1\Rightarrow x=1\Rightarrow y=\cfrac{9}{4}$