Question

Решить систему!;50 баллов!
$\sqrt{x} + \sqrt{y} = 3 \\ \sqrt{x + 5} + \sqrt{y + 3} = 5$
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Answer 1

$\displaystyle\begin{cases}\sqrt x+\sqrt y=3\\\sqrt{x+5}+\sqrt{y+3}=5\end{cases}\\\sqrt{x}+\sqrt{y}=3\\(\sqrt{x}+\sqrt{y})^2=9\\x+y+2\sqrt{xy}=9\\9-(x+y)=2\sqrt{xy}\\(9-(x+y))^2=4xy\\\\\sqrt{x+5}+\sqrt{y+3}=5\\(\sqrt{x+5}+\sqrt{y+3})^2=25\\x+5+y+3+2\sqrt{(x+5)(y+3)}=25\\(17-(x+y))^2=4(x+5)(y+3)\\(17-(x+y))^2=4(xy+3x+5y+15)\\(17-(x+y))^2=4xy+4(3x+5y+15)\\(17-(x+y))^2=(9-(x+y))^2+4(3x+5y+15)\\(17-(x+y))^2-(9-(x+y))^2=4(3x+5y+15)\\(17-(x+y)-9+(x+y))(17-(x+y)+9-(x+y))=4(3x+5y+15)\\8(26-2(x+y))=4(3x+5y+15)$

$\displaystyle 2(26-2(x+y))=3x+5y+15\\52-4x-4y=3x+5y+15\\7x+9y=37\\x=\frac{37-9y}{7}\\(9-(x+y))^2=4xy\\(9-(\frac{37-9y}{7}+y))^2=4y(\frac{37-9y}{7})\\81-18(\frac{37-2y}{7})+(\frac{37-2y}{7})^2=4y(\frac{37-9y}{7})\\81*49-126(37-2y)+(37-2y)^2=28y(37-9y)\\81*49-126*37+252y+37^2-4*37y+4y^2=28*37y-28*9y^2\\256y^2-932y+676=0|:4\\64y^2-233y+169=0\\y_{1,2}=\frac{233\pm\sqrt{233^2-4*64*169}}{128}=\frac{233\pm105}{128}$

$\displaystyle y_1=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y_2=\frac{338}{128}=\frac{169}{64}=2.640625\\x_1=\frac{37-9}{7}=4\ \ \ x_2=\frac{37-9*\frac{169}{64}}{7}=\frac{37-23\frac{49}{64}}{7}=\frac{847}{448}=1.890625$

$OTBET:(4;1);(1.890625;2.640625)$

Answer 2

$\begin{cases}\sqrt x+\sqrt y=3\\\sqrt{x+5}+\sqrt{y+3}=5\end{cases} \\\\\\\sqrt{x}+\sqrt{y}=3\\\\\sqrt{x}=3-\sqrt{y}\\\\(\sqrt{x})^2=(3-\sqrt{y})^2\\\\x=9-6\sqrt{y}+y$

.

$\sqrt{9-6\sqrt{y}+y+5}+\sqrt{y+3}=5\\\\\sqrt{14-6\sqrt{y}+y}=5-\sqrt{y+3}\\\\\Big(\sqrt{14-6\sqrt{y}+y} \Big)^2=\Big(5-\sqrt{y+3} \Big)^2\\\\14-6\sqrt{y}+y=25-10\sqrt{y+3}+y+3\\\\-6\sqrt{y}=28-10\sqrt{y+3}+3-14\\\\-6\sqrt{y}=14-10\sqrt{y+3}\\\\\Big(-6\sqrt{y} \Big)^2=\Big(14-10\sqrt{y+3} \Big)^2\\\\36y=196-280\sqrt{y+3}+100y+300\\\\280\sqrt{y+3}=496+64y \ \ \ | \ \ \div8\\\\35\sqrt{y+3}=62+8y\\\\\Big(35\sqrt{y+3}\Big)^2=\big(62+8y\big)^2\\\\1225y+3675=3844+992y+64y^2$

$64y^2-233y+169=0\\\\64y(y-1)-169(y-1)=0\\\\(y-1)(64y-169)=0\\\\y_1=1 \ \ \ \ y_2=\frac{169}{64}$

.

$x_1=9-6\sqrt{1}+1=4\\\\x_2=9-6\sqrt{\frac{169}{64}}+\frac{169}{64}=\frac{121}{64}\\\\\\\boxed{(4; \; 1) \; ; \; \bigg(\frac{121}{64} \; ; \; \frac{169}{64} \bigg)}$