Question

Решите систему !$3x+5yx^{2} =20\\ xy^{2} =5$ дам !100!

Answer 1

$\left\{\begin{array}{ccc}3x+5y^{2}=20 \\xy^{2}=5 \end{array}\right\\\\\\\left\{\begin{array}{ccc}5y^{2}=20-3x \\xy^{2}=5 \end{array}\right\\\\\\\left\{\begin{array}{ccc}y^{2}=4-0,6x \\x\cdot(4-0,6x)=5 \end{array}\right\\\\\\\left\{\begin{array}{ccc}y^{2}=4-0,6x \\4x-0,6x^{2}- 5=0 \ |\cdot(-5) \end{array}\right\\\\\\\left\{\begin{array}{ccc}y^{2}=4-0,6x \\3x^{2}-20x+25=0 \end{array}\right\\\\\\3x^{2}-20x+25=0\\\\D=(-20)^{2}-4\cdot 3\cdot 25=400-300=100=10^{2}$

$x_{1}=\dfrac{20-10}{6}=\dfrac{10}{6}=\dfrac{5}{3}=1\dfrac{2}{3}\\\\x_{2}=\dfrac{20+10}{6}=\dfrac{30}{6} =5\\\\y^{2} =4-0,6\cdot\frac{5}{3}=4-\dfrac{6}{10}\cdot \dfrac{5}{3}=4-1=3\\\\y_{1,2}=\pm\sqrt{3} \\\\y^{2}=4-0,6\cdot 5=4-3=1\\\\y_{3,4}=\pm 1\\\\Otvet;\boxed{(1\dfrac{2}{3} \ ; \ -\sqrt{3}),(1\dfrac{2}{3} \ ; \ \sqrt{3}),(5 \ ; \ -1),(5 \ ; \ 1) }$