А) 9x^4-13x^2+4=0
пусть x^2 = t, тогда
9t^2-13t+4=0
D=169-4*36=25
t1=1
t2=4/9
если t=1,то х= плюс минус 1
если t=4/9,то х= плюс минус 2/3
1.ОДЗ:xєR
2.f'(x)=12x+3x^2
3.12x+3x^2=04x+x^2
x(4+x)=0
x=0 или 4+x=0 x=-4 не положено [-3;2]f(-3)=3*(-3)^4+(-3)^3+7=227f(0)=7f(2)=31maxf(x)=f(-3)=227minf(x)=f(0)=7
![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