А) -12х + 5х -4х = -11x
б) b - 6а -10b + 9а + 4b = 3a - 5b
5^7 * 5^4 * 4^5 = 5^11 * 4^5 = 48828125 * 1024 = 50 000 000 000
Существование корней:
![D=(-2(a+3))^2-4\cdot (a-1)\cdot 2a=4(a^2+6a+9)-8a(a-1)=\\ =4a^2+24a+36-8a^2+8a=-4a^2+32a+36>0\\ \\ a^2-8a-9<0\\ (a-4)^2-25<0\\ |a-4|<5\\ -5<a-4<5\\ -1<a<9](https://tex.z-dn.net/?f=D%3D%28-2%28a%2B3%29%29%5E2-4%5Ccdot+%28a-1%29%5Ccdot+2a%3D4%28a%5E2%2B6a%2B9%29-8a%28a-1%29%3D%5C%5C+%3D4a%5E2%2B24a%2B36-8a%5E2%2B8a%3D-4a%5E2%2B32a%2B36%3E0%5C%5C+%5C%5C+a%5E2-8a-9%3C0%5C%5C+%28a-4%29%5E2-25%3C0%5C%5C+%7Ca-4%7C%3C5%5C%5C+-5%3Ca-4%3C5%5C%5C+-1%3Ca%3C9)
Квадратное уравнение имеет два различных положительных корня, если
![\displaystyle \left \{ {{\dfrac{2(a+3)}{a-1}>0} \atop {\dfrac{2a}{a-1}>0}} \right. ~~~\Leftrightarrow~~~\left \{ {{a \in (-\infty;-3)\cup (1;+\infty)} \atop {a\in (-\infty;0)\cup(1;+\infty)}} \right. ~~\Leftrightarrow~~~ \left[\begin{array}{ccc}a<-3\\ \\ a>1\end{array}\right](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cleft+%5C%7B+%7B%7B%5Cdfrac%7B2%28a%2B3%29%7D%7Ba-1%7D%3E0%7D+%5Catop+%7B%5Cdfrac%7B2a%7D%7Ba-1%7D%3E0%7D%7D+%5Cright.+~~~%5CLeftrightarrow~~~%5Cleft+%5C%7B+%7B%7Ba+%5Cin+%28-%5Cinfty%3B-3%29%5Ccup+%281%3B%2B%5Cinfty%29%7D+%5Catop+%7Ba%5Cin+%28-%5Cinfty%3B0%29%5Ccup%281%3B%2B%5Cinfty%29%7D%7D+%5Cright.+~~%5CLeftrightarrow~~~+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%3C-3%5C%5C+%5C%5C+a%3E1%5Cend%7Barray%7D%5Cright)
Общее решение: ![a \in (1;9)](https://tex.z-dn.net/?f=a+%5Cin+%281%3B9%29)
Ответ: ![a \in (1;9)](https://tex.z-dn.net/?f=a+%5Cin+%281%3B9%29)
0,3^-2+(3/7)^-1+(-0,5)^-2*(3/4)+(-1)^-8*6=100/9+7/3+4*(3/4)+6=100/9+7/3+9=22 целых 4/9
(2/3)^-2-(1/9)^-1+(6/17)^0*(1/8)-0,25^-2*16=9/4-9+1/8-16*16=19/8-265=-262 целых 3/8