Question

Решите систему уравнений

Answer 1

Ответ:

$(28; 16\sqrt{3});$

Объяснение:

ОДЗ:

$\left \{ {{x+y \geq 0} \atop {x-y \geq 0}} \right. \Leftrightarrow \left \{ {{x \geq -y} \atop {x \geq y}} \right. \Leftrightarrow x \geq y;$

Решение:

$\left \{ {{\sqrt[4]{x+y}-\sqrt[4]{x-y}=2}; \atop {\sqrt{x+y}+\sqrt{x-y}=8}.} \right. \Leftrightarrow \left \{ {{(\sqrt[4]{x+y}-\sqrt[4]{x-y})^{2}=2^{2}}; \atop {\sqrt{x+y}+\sqrt{x-y}=8}.} \right. \Leftrightarrow \left \{ {{\sqrt{x+y}-2\sqrt[4]{x^{2}-y^{2}}+\sqrt{x-y}=4}; \atop {\sqrt{x+y}+\sqrt{x-y}=8}.} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{8-2\sqrt[4]{x^{2}-y^{2}}=4}; \atop {\sqrt{x+y}+\sqrt{x-y}=8}.} \right. \Leftrightarrow \left \{ {{2\sqrt[4]{x^{2}-y^{2}}=4}; \atop {(\sqrt{x+y}+\sqrt{x-y})^{2}=8^{2}}.} \right. \Leftrightarrow \left \{ {{\sqrt[4]{x^{2}-y^{2}}=2}; \atop {x+y+2\sqrt{x^{2}-y^{2}}+x-y=64}.} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{(\sqrt[4]{x^{2}-y^{2}})^{4}=2^{4}}; \atop {2x+2\sqrt{x^{2}-y^{2}}=64}.} \right. \Leftrightarrow \left \{ {{x^{2}-y^{2}=16}; \atop {\sqrt{x^{2}-y^{2}}=32-x}.} \right. \Leftrightarrow \left \{ {{x^{2}-y^{2}=16}; \atop {(\sqrt{x^{2}-y^{2}})^{2}=(32-x)^{2}}.} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x^{2}-y^{2}=16}; \atop {x^{2}-y^{2}=1024-64x+x^{2}}.} \right. \Leftrightarrow \left \{ {{y^{2}=x^{2}-16}; \atop {x^{2}-64x+1024=16}.} \right. \Leftrightarrow \left \{ {{y= \pm \sqrt{x^{2}-16}}; \atop {x^{2}-64x+1008=0}.} \right.$

$x^{2}-64x+1008=0;$

$D=b^{2}-4ac;$

$D=(-64)^{2}-4 \cdot 1 \cdot 1008=64^{2}-4032=(60+4)^{2}-4032=60^{2}+2 \cdot 60 \cdot 4+4^{2}-$

$-4032=3600+480+16-4032=4096-4032=64;$

$x_{1,2}=\frac{-b \pm \sqrt{D}}{2a};$

$x_{1}=\frac{-(-64)+\sqrt{64}}{2 \cdot 1}=\frac{64+8}{2}=\frac{64}{2}+\frac{8}{2}=32+4=36;$

$x_{2}=\frac{-(-64)-\sqrt{64}}{2 \cdot 1}=\frac{64-8}{2}=\frac{64}{2}-\frac{8}{2}=32-4=28;$

$y= \pm \sqrt{x^{2}-16};$

$y_{1,2}= \pm \sqrt{36^{2}-16}= \pm \sqrt{1296-16}= \pm \sqrt{1280}= \pm \sqrt{256 \cdot 5}= \pm 16\sqrt{5};$

$y_{3,4}= \pm \sqrt{28^{2}-16}= \pm \sqrt{784-16}= \pm \sqrt{768}= \pm \sqrt{256 \cdot 3}= \pm 16\sqrt{3};$

$28<16\sqrt{5};$

$28-16\sqrt{3}<28+16\sqrt{3};$

$(28; 16\sqrt{3});$