6x-3(3x-2)=18
6x-9x+6=18
-3x+6=18
-3x=18-6
-3x=12|*(-3)
x=-4
Смотри формула для всех одна
D=b^2-4*a*c
дальше находишь два члена (либо x1 и x2, либо y1 и y2)
ФОРМУЛА ДЛЯ НИХ : x1x2=(-b+-корень из Д)/ 2a
пример под буквой а : Д=-9+4*5*8= 169
х1=1 х2=-1,6
посмотри моё решение,может тебе поможет.Я сама на 100\% не уверена
Составляем систему:
![\left \{ {{ \frac{x^2+2x+4}{x^2-2x-2} \leq 1 } \atop {\frac{x^2+2x+4}{x^2-2x-2} \geq -1}} \right. \Rightarrow \left \{ {{ \frac{x^2+2x+4-x^2+2x+2}{x^2-2x-2} \leq 0} \atop { \frac{x^2+2x+4+x^2-2x-2}{x^2-2x-2} \geq 0 }} \right. \Rightarrow \left \{ {{ \frac{4x+6}{x^2-2x-2} \leq 0 } \atop { \frac{2x^2+2}{x^2-2x-2} \geq 0 }} \right. ](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B+%5Cfrac%7Bx%5E2%2B2x%2B4%7D%7Bx%5E2-2x-2%7D+%5Cleq+1+%7D+%5Catop+%7B%5Cfrac%7Bx%5E2%2B2x%2B4%7D%7Bx%5E2-2x-2%7D++%5Cgeq+-1%7D%7D+%5Cright.+%5CRightarrow++%5Cleft+%5C%7B+%7B%7B+%5Cfrac%7Bx%5E2%2B2x%2B4-x%5E2%2B2x%2B2%7D%7Bx%5E2-2x-2%7D++%5Cleq+0%7D+%5Catop+%7B+%5Cfrac%7Bx%5E2%2B2x%2B4%2Bx%5E2-2x-2%7D%7Bx%5E2-2x-2%7D+%5Cgeq+0+%7D%7D+%5Cright.+%5CRightarrow++%5Cleft+%5C%7B+%7B%7B+%5Cfrac%7B4x%2B6%7D%7Bx%5E2-2x-2%7D+%5Cleq+0+%7D+%5Catop+%7B+%5Cfrac%7B2x%5E2%2B2%7D%7Bx%5E2-2x-2%7D+%5Cgeq+0+%7D%7D+%5Cright.+%0A)
решаем каждое неравенство по отдельности:
![\frac{4x+6}{x^2-2x-2} \leq 0 \\ \frac{x+3}{x^2-2x-2} \leq 0 \\x^2-2x-2=0 \\D=4+8=12=(2\sqrt{3})^2 \\x_1= \frac{2+2\sqrt{3}}{2} =1+\sqrt{3} \\x_2=1-\sqrt{3} \\\frac{2x+3}{x^2-2x-2} \leq 0](https://tex.z-dn.net/?f=%5Cfrac%7B4x%2B6%7D%7Bx%5E2-2x-2%7D+%5Cleq+0%0A%5C%5C+%5Cfrac%7Bx%2B3%7D%7Bx%5E2-2x-2%7D++%5Cleq+0%0A%5C%5Cx%5E2-2x-2%3D0%0A%5C%5CD%3D4%2B8%3D12%3D%282%5Csqrt%7B3%7D%29%5E2%0A%5C%5Cx_1%3D+%5Cfrac%7B2%2B2%5Csqrt%7B3%7D%7D%7B2%7D+%3D1%2B%5Csqrt%7B3%7D%0A%5C%5Cx_2%3D1-%5Csqrt%7B3%7D%0A%5C%5C%5Cfrac%7B2x%2B3%7D%7Bx%5E2-2x-2%7D++%5Cleq+0)
используем метод интервалов(см. приложение 1)
![x \in (-\infty;- 1,5]\cup (1-\sqrt{3};1+\sqrt{3})](https://tex.z-dn.net/?f=x+%5Cin+%28-%5Cinfty%3B-+1%2C5%5D%5Ccup+%281-%5Csqrt%7B3%7D%3B1%2B%5Csqrt%7B3%7D%29)
решаем 2 неравенство:
![\frac{2x^2+2}{x^2-2x-2} \geq 0 \\\frac{x^2+1}{x^2-2x-2} \geq 0 \\x^2+1\ \textgreater \ 0,\ \forall \ x \in R \\ \frac{1}{x^2-2x-2} \geq 0 \\x^2-2x-2\ \textgreater \ 0 \\x_1=1+\sqrt{3} \\x_2=1-\sqrt{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2x%5E2%2B2%7D%7Bx%5E2-2x-2%7D+%5Cgeq+0%0A%5C%5C%5Cfrac%7Bx%5E2%2B1%7D%7Bx%5E2-2x-2%7D+%5Cgeq+0%0A%5C%5Cx%5E2%2B1%5C+%5Ctextgreater+%5C+0%2C%5C+%5Cforall+%5C+x+%5Cin+R%0A%5C%5C+%5Cfrac%7B1%7D%7Bx%5E2-2x-2%7D++%5Cgeq++0%0A%5C%5Cx%5E2-2x-2%5C+%5Ctextgreater+%5C+0%0A%5C%5Cx_1%3D1%2B%5Csqrt%7B3%7D%0A%5C%5Cx_2%3D1-%5Csqrt%7B3%7D)
используем метод интервалов(см. приложение 2)
![x \in (-\infty;1-\sqrt{3})\cup (1+\sqrt{3};+\infty)](https://tex.z-dn.net/?f=x+%5Cin+%28-%5Cinfty%3B1-%5Csqrt%7B3%7D%29%5Ccup+%281%2B%5Csqrt%7B3%7D%3B%2B%5Cinfty%29)
пересекаем множества решений этих двух неравенств:
![x \in ((-\infty;- 1,5]\cup (1-\sqrt{3};1+\sqrt{3}))\cap ((-\infty;1-\sqrt{3})\cup (1+\sqrt{3};+\infty))\\=(-\infty;-1,5]](https://tex.z-dn.net/?f=x+%5Cin+%28%28-%5Cinfty%3B-+1%2C5%5D%5Ccup+%281-%5Csqrt%7B3%7D%3B1%2B%5Csqrt%7B3%7D%29%29%5Ccap++%28%28-%5Cinfty%3B1-%5Csqrt%7B3%7D%29%5Ccup+%281%2B%5Csqrt%7B3%7D%3B%2B%5Cinfty%29%29%5C%5C%3D%28-%5Cinfty%3B-1%2C5%5D)
наибольшее целое отрицательное: -2
Ответ: -2
<span>a) 5(x+1.4y)-0.8(2x+y)=5х+7у-1,6х-0,8у=3,4х+6,2
</span><span>б)2/3(x-y-z)-(2/3x-y-z)=2/3х-2/3у-2/3z-2/3x+y+z=1/3y+1/3z=1/3(y+z)
</span><span>в)-a+0.5(3a+0.2b)-(a+0.1b)=-a+1.5a+0.1b-a-0.1b=-0.5a
</span><span>г)-10(2/5b+1/2)+3/4(8-b)+5b=4b-5+6-3/4b=3.25b+1</span>