(2^(2/3) *а^(2/3) *а^(1/2) *b^(1/2))/(a*b^(1/3)*2^(1/2) = 2^(1/6) *а^(1/6)*b^(1/6)= √2аb - здесь корень шестой степени
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)
Y=-2/3*x+6
x=0 y=6
y=0 -2/3*x+6=0⇒2/3*x=6⇒x=6:2/3=6*3/2=9
Ответ (0;6);(9;0)
<span>2sin 2n/3- ctg n/6 =2*(√3/2) - √3 = √3 - √3 = 0</span>
Ответ на пример
: = a63 + b63