Question

Решите системы уравнений и подробно распишите решение​

Answer 1

1) $\left \{ {{y=0,5x^2} \atop {y-x=1}} \right.$

0,5x^2 - x = 1

x=1 + $\sqrt{3}$

x = 1 - $\sqrt{3}$

y= 0,5(1+$\sqrt{3}$)^2

y=0,5(1-$\sqrt{3}$)^2

y=2+ $\sqrt{3}$

y=2-$\sqrt{3}$

($x_{1\\}$, $y_{1}$) = (1+$\sqrt{3}$, 2+$\sqrt{3}$)

($x_{2}$, $y_{2}$) = (1-$\sqrt{3}$, 2-$\sqrt{3}$)

$\left \{ {{2+\sqrt{3}= 0,5(1+\sqrt{3}) ^2 } \atop 2+ \sqrt{3} - (1+ \sqrt{3} ) = 1$

$\left \{ {{2-\sqrt{3}=0,5 (1+\sqrt{3})^2 } \atop {2-\sqrt{3}-(1-\sqrt{3})=1 }} \right.$

$\left \{ {{3,73205=3,73205} \atop {1=1}} \right.$

$\left \{ {{0,267949=0,267949} \atop {1=1}} \right.$

$(x_{1} , y_{1} ) = (1 + \sqrt{3} , 2+ \sqrt{3} )\\$

$( x_{2} , y_{2} ) = (1- \sqrt{3} , 2-\sqrt{3} )$

2) $\left \{ {{x+2y=1} \atop {2x+y^2= -1}} \right.$

$\left \{ {{x=1-2y} \atop {2x+y^2=-1}} \right.$

$2(1-2y) +y^2 = - 1$

$y=1\\y=3$

$x=1-2*1\\x=1-2*3$

$x=-1\\x=-5$

$(x_{1} , y_{1} ) = (-1,1) \\( x_{2} , y_{2} = (-5,3)$

$\left \{ {{-1+2*1=1} \atop {x*(-1)+1^2=-1}} \right.$

$\left \{ {{-5+2*3=1} \atop {2*(-5)+3^2 = -1}} \right.$

$\left \{ {{1=1} \atop {-1=-1}} \right.$

$\left \{ {{1=1} \atop {-1=-1}} \right.$

$( x_{1} , y_{1} ) = (-1,1)\\( x_{2} , y_{2} ) = (-5,3)$

3)  $\left \{ {{x^{2} +xy-y^2 = 4} \atop {3x+y=10}} \right.$

$\left \{ {{x^{2} + xy-y^2=4} \atop {y=9}} \right.$

$x^2+x*9-9^2=4$

$x= \frac{-9+\sqrt{421} }{2} \\x= \frac{-9-\sqrt{421} }{2}$

$( x_{1} , y_{1} ) = ( \frac{-9+\sqrt{421} }{2} , 9 )\\( x_{2} , y_{2} ) = ( \frac{-9-\sqrt{421} }{2} , 9)$

$\left \{ {{(\frac{-9+\sqrt{421} }{2})^2+\frac{-9+\sqrt{421} }{2}*9-9^2=4) } \atop {1+9=10}} \right.$

$\left \{ {{( \frac{-9-\sqrt{421} }{2})^2+\frac{-9-\sqrt{421} }{2}*9-9^2=4 } \atop {1+9=10}} \right.$

$\left \{ {{4=4} \atop {10=10}} \right.$

$\left \{ {{4=4} \atop {10=10}} \right.$

$( x_{1} , y_{1} ) = ( \frac{-9+\sqrt{421} }{2} , 9)\\( x_{2} , y_{2} ) = ( \frac{-9-\sqrt{421} }{2} , 9 )$

4)  $\left \{ {{\frac{1}{x}-\frac{1}{y}=\frac{1}{12} } \atop {2x-y=2}} \right.$

$\left \{ {{\frac{1}{x}-\frac{1}{y}=\frac{1}{12} } \atop {y=-2+2x}} \right.$

$\frac{1}{x}-\frac{1}{-2+2x} = \frac{1}{12}$

$x=3\\x=4$

$y=-2+2*3\\y=-2+2*4$

$y=4\\y=6$

$( x_{1} , y_{1} ) = (3,4)\\( x_{2} , y_{2} ) = (4,6)$