Question

Ссистемой уравнений

Answer 1

$\left \{ {{12(x+y)^{2}+x=2,5-y } \atop {6(x-y)^{2}+x }=0,125+y} \right. \\\\\left \{ {{12(x+y)^{2} +(x+y)-2,5=0} \atop {6(x-y)^{2}+(x-y)-0,125=0 }} \right.$

1) Обозначим x + y = m , тогда :

12m² + m - 2,5 = 0

D = 1² - 4 * 12 * (- 2,5) = 1 + 120 = 121 = 11²

$m_{1}=\frac{-1+11}{24}=\frac{10}{24}=\frac{5}{12}\\\\m_{2} =\frac{-1-11}{24}=-\frac{12}{24}=-\frac{1}{2}\\\\\left[\begin{array}{ccc}x+y=\frac{5}{12} \\x+y=-\frac{1}{2}\end{array}\right$

2) Обозначим x - y = n , тогда :

6n² + n - 0,125 = 0

D = 1² - 4 * 6 * (- 0,125) = 1 + 3 = 4 = 2²

$n_{1}=\frac{-1+2}{12}=\frac{1}{12}\\\\n_{2}=\frac{-1-2}{12}=-\frac{3}{12}=-\frac{1}{4}\\\\\left[\begin{array}{ccc}x-y=\frac{1}{12} \\x-y=-\frac{1}{4} \end{array}\right$

Получим 4 системы уравнений :

1)

$+\left \{ {{x+y=\frac{5}{12} } \atop {x-y=\frac{1}{12} }} \right.\\ ------\\2x=\frac{1}{2}\\x=\frac{1}{4}\\y=\frac{5}{12}-\frac{1}{4}=\frac{1}{6}$

2)

$+\left \{ {{x+y=\frac{5}{12} } \atop {x-y=-\frac{1}{4} }} \right. \\------\\2x=\frac{1}{6}\\x=\frac{1}{12}\\y=\frac{5}{12}-\frac{1}{12}=\frac{1}{3}$

3)

$+\left \{ {{x+y=-\frac{1}{2} } \atop {x-y=\frac{1}{12} }} \right.\\ ------\\2x=-\frac{5}{12}\\x=-\frac{5}{24}\\y=-\frac{1}{2}+\frac{5}{24}=-\frac{7}{24}$

4)

$+\left \{ {{x+y=-\frac{1}{2} } \atop {x-y=-\frac{1}{4} }} \right.\\ ------\\2x=-\frac{3}{4}\\x=-\frac{3}{8}\\y=-\frac{1}{2}+\frac{3}{8}=-\frac{1}{8}$

$Otvet:(\frac{1}{4};\frac{1}{6}),(\frac{1}{12};\frac{1}{3}),(-\frac{5}{24};-\frac{7}{24}),(-\frac{3}{8};-\frac{1}{8})$