Question

алгебра 9 класс срочно даю 35 баллов!

Answer 1

а)

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

Попробуем подставить x=1 и y=2

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

$\displaystyle \left \{ {{1=1} \atop {2=2}} \right.$

Эта пара чисел подходит

б)

⁻$\displaystyle \left \{ {3x-11y=36} \atop {3x+11y=-30} \right.$

$\displaystyle -11y-11y=36-(-30)$

$\displaystyle -22y=66$

$\displaystyle y=-3$

$\displaystyle x=\frac{36+11*(-3)}{3} =\frac{3*(12-11)}{3} =1$

Ответ: (1;-3)

в)

$\displaystyle \left \{ {{x-2y=-7} \atop {3x+4y=19}} \right.$

$\displaystyle \left \{ {{x=2y-7} \atop {3*(2y-7)+4y=19}} \right.$

$\displaystyle \left \{ {{x=2y-7} \atop {6y-21+4y=19}} \right.$

$\displaystyle \left \{ {{x=2y-7} \atop {10y=40}} \right.$

$\displaystyle \left \{ {{x=2*4-7} \atop {y=4}} \right.$

$\displaystyle \left \{ {{x=1} \atop {y=4}} \right.$

Ответ: (1;4)

г)

$\displaystyle \left \{ {{(x-y)^{2}-x+y=0} \atop {x^{2}y^{2}-xy-2=0}} \right.$

$\displaystyle \left \{ {{(x-y)^{2}-(x-y)=0} \atop {x^{2}y^{2}-xy-2=0}} \right.$

Пусть x-y=a, тогда xy=b

$\displaystyle \left \{ {{a^{2}-a=0} \atop {b^{2}-b-2=0}} \right.$

$\displaystyle a^{2}-a=0$

$\displaystyle a*(a-1)=0$

$\displaystyle a=0 ; a=1$

$\displaystyle b^{2}-b-2=0$

$\displaystyle D = (-1)^{2}-4*1*(-2)=1+8=9=3^{2}$

$\displaystyle b_{1} =\frac{1+3}{2*1}= \frac{4}{2} =2$

$\displaystyle b_{2} =\frac{1-3}{2*1}= -\frac{2}{2} =-1$

Вернёмся к замене:

Составим 4 системы:

1)$\displaystyle \left \{ {{x-y=0} \atop {xy=2}} \right.$

$\displaystyle \left \{ {{x=y} \atop {y^{2}=2}} \right.$

$\displaystyle \left \{ {{x=\sqrt{2} } \atop {y=\sqrt{2} }} \right.$

2)$\displaystyle \left \{ {{x-y=0} \atop {xy=-1}} \right.$

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

нет решений

3)$\displaystyle \left \{ {{x-y=1} \atop {xy=2}} \right.$

$\displaystyle \left \{ {{x=1+y} \atop {(1+y)y=2}} \right.$

$\displaystyle \left \{ {{x=1+y} \atop {y^{2}+y-2=0}} \right.$

$\displaystyle y^{2}+y-2=0$

$\displaystyle D = 1^{2}-4*1*(-2)=1+8=9=3^{2}$

$\displaystyle y_{1}=\frac{-1+3}{2*1}=\frac{2}{2}=1$

$\displaystyle y_{2}=\frac{-1-3}{2*1}=--\frac{4}{2}=-2$

$\displaystyle x_{1} =1+1=2$

$\displaystyle x_{2} =1+(-2)=-1$

4)$\displaystyle \left \{ {{x-y=1} \atop {xy=-1}} \right.$

$\displaystyle \left \{ {{x=1+y} \atop {(1+y)y=-1}} \right.$

$\displaystyle \left \{ {{x=1+y} \atop {y^{2}+y+1=0}} \right.$

$\displaystyle y^{2}+y+1=0$

$\displaystyle D = 1^{2}-4*1*1=1-4=-3}$

Нет решений

Ответ: (√2;√2),(2;1),(-1;-2)