(а-1/a^2)*(ax-a/a-1)+(x-1/2a)=((a-1)*a(x-1)/a^2*(a-1))+((1-x)/2a)=((x-1)/a)+((1-x)/2a)=
=((2x-2)/2a)+((1-x)/2a)=((2x-2+1-x)/2a)=x-1/2a
"/"-дробь
4(4с-3)-(10с+8)=16с-12-10с-8=6с-20 при с= 5/6, получим
6*5/6-20=5-20=-15.
Ответ:-15
удачи!!!
Ответ будет такой:
8х-4у-9х-3у
-х-4у-3у
Ответ: -х-7у
![a^2+2b^2+2ab+2b+2\ \textgreater \ 0\\\\ a^2+2ab+b^2+b^2+2b+1+1\ \textgreater \ 0\\\\ (a+b)^2+(b+1)^2+1\ \textgreater \ 0\\\\](https://tex.z-dn.net/?f=a%5E2%2B2b%5E2%2B2ab%2B2b%2B2%5C+%5Ctextgreater+%5C+0%5C%5C%5C%5C%0Aa%5E2%2B2ab%2Bb%5E2%2Bb%5E2%2B2b%2B1%2B1%5C+%5Ctextgreater+%5C+0%5C%5C%5C%5C%0A%28a%2Bb%29%5E2%2B%28b%2B1%29%5E2%2B1%5C+%5Ctextgreater+%5C+0%5C%5C%5C%5C)
![(a+b)^2 \geq 0](https://tex.z-dn.net/?f=%28a%2Bb%29%5E2+%5Cgeq+0)
и также
![(b+1)^2 \geq 0](https://tex.z-dn.net/?f=%28b%2B1%29%5E2+%5Cgeq+0)
по этому
![(a+b)^2+(b+1)^2+1\ \textgreater \ 0](https://tex.z-dn.net/?f=%28a%2Bb%29%5E2%2B%28b%2B1%29%5E2%2B1%5C+%5Ctextgreater+%5C+0)
вне зависимости от значений
![a](https://tex.z-dn.net/?f=a)
и
![b](https://tex.z-dn.net/?f=b)
Ответ:
![a\in(-\infty;\ +\infty)](https://tex.z-dn.net/?f=a%5Cin%28-%5Cinfty%3B%5C+%2B%5Cinfty%29)
и