2(х²-6х+9)-4=2х²-12х+18-4=2х²-12х+14
ax²+bx+c - парабола
вершина вычисляется по формуле
![- \frac{b}{2a} = \frac{12}{2 \times 2} = 3](https://tex.z-dn.net/?f=+-++%5Cfrac%7Bb%7D%7B2a%7D++%3D++%5Cfrac%7B12%7D%7B2+%5Ctimes+2%7D++%3D+3)
<span>(32х^-10)^-3/5 = 2^5*(-3/5)* x^-10*(-3/5) = 1/8 * x^6</span>
![1)\ sin\ x+cos\ x=1\ \ |* \frac{ \sqrt{2} }{2} \\ \frac{ \sqrt{2} }{2}\ sin\ x+\frac{ \sqrt{2} }{2}\ cos\ x=\frac{ \sqrt{2} }{2}\\ \ cos\ \frac{\pi}{4}\ sin\ x +sin\ \frac{\pi}{4}\ cos\ x=\frac{ \sqrt{2} }{2}\\ (P.\ S.\ sin\ (a+b)=sin\ a\ cos\ b+ cos\ a\ sin\ b)\\ sin\ ( \frac{\pi}{4}+x)=\frac{ \sqrt{2} }{2}\\ ](https://tex.z-dn.net/?f=1%29%5C+sin%5C+x%2Bcos%5C+x%3D1%5C+%5C+%7C%2A+%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D+%5C%5C%0A%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D%5C+sin%5C+x%2B%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D%5C+cos%5C+x%3D%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D%5C%5C%0A%5C+cos%5C+%5Cfrac%7B%5Cpi%7D%7B4%7D%5C+sin%5C+x+%2Bsin%5C++%5Cfrac%7B%5Cpi%7D%7B4%7D%5C+cos%5C+x%3D%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D%5C%5C%0A%28P.%5C+S.%5C+sin%5C+%28a%2Bb%29%3Dsin%5C+a%5C+cos%5C+b%2B+cos%5C+a%5C+sin%5C+b%29%5C%5C%0Asin%5C+%28+%5Cfrac%7B%5Cpi%7D%7B4%7D%2Bx%29%3D%5Cfrac%7B+%5Csqrt%7B2%7D+%7D%7B2%7D%5C%5C%0A)
![\left[\begin{gathered}\ \frac{\pi}{4}+x=\frac{\pi}{4}+2\pi n ,\hfill \\ \frac{\pi}{4}+x=\pi-\frac{\pi}{4}+2\pi n,\ n\in \mathcal{Z};\ \ \ \end{gathered} \left[\begin{gathered}\ x=2\pi n,\hfill \\ x= \frac{\pi}{2}+2\pi n,\ n\in \mathcal Z. \\ \end{gathered} ](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Bgathered%7D%5C%0A+%5Cfrac%7B%5Cpi%7D%7B4%7D%2Bx%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%2B2%5Cpi+n++%2C%5Chfill+%5C%5C+%0A%5Cfrac%7B%5Cpi%7D%7B4%7D%2Bx%3D%5Cpi-%5Cfrac%7B%5Cpi%7D%7B4%7D%2B2%5Cpi+n%2C%5C+n%5Cin+%5Cmathcal%7BZ%7D%3B%5C+%5C+%5C++%5Cend%7Bgathered%7D+%0A%5Cleft%5B%5Cbegin%7Bgathered%7D%5C%0Ax%3D2%5Cpi+n%2C%5Chfill+%5C%5C+%0Ax%3D+%5Cfrac%7B%5Cpi%7D%7B2%7D%2B2%5Cpi+n%2C%5C+n%5Cin+%5Cmathcal+Z.++%5C%5C+%5Cend%7Bgathered%7D+%0A)
На отрезке [-π; 2π]:
0, π/2, 2π, то есть
три корня.![2)\ tg\ x+ \frac{1}{tg\ x}=2\\ OD3: \left \{ {{\ tg\ x\neq0,} \atop {cos\ x\neq 0; }} \right. \ \ \left \{ {{x\neq \pi n,} \atop {x\neq \dfrac{\pi}{2}+\pi n.}} \right. \Rightarrow x\neq \frac{\pi n}{2},\ n\in \mathcal Z\\ tg\ x+ \frac{1}{tg\ x}-2=0\\ \frac{tg^2x+1-2tg\ x}{tg\ x}=0\\ tg^2x-2tg\ x+1=0\\ D=4-4=0\\ tg\ x= \frac{2}{2}=1\\ x= \frac{\pi}{4}+\pi n, n\in \mathcal Z](https://tex.z-dn.net/?f=2%29%5C+tg%5C+x%2B+%5Cfrac%7B1%7D%7Btg%5C+x%7D%3D2%5C%5C%0AOD3%3A+%5Cleft+%5C%7B+%7B%7B%5C+tg%5C+x%5Cneq0%2C%7D+%5Catop+%7Bcos%5C+x%5Cneq++0%3B+%7D%7D+%5Cright.+%5C+%5C++%5Cleft+%5C%7B+%7B%7Bx%5Cneq+%5Cpi+n%2C%7D+%5Catop+%7Bx%5Cneq+%5Cdfrac%7B%5Cpi%7D%7B2%7D%2B%5Cpi+n.%7D%7D+%5Cright.+%5CRightarrow+x%5Cneq++%5Cfrac%7B%5Cpi+n%7D%7B2%7D%2C%5C+n%5Cin+%5Cmathcal+Z%5C%5C%0Atg%5C+x%2B+%5Cfrac%7B1%7D%7Btg%5C+x%7D-2%3D0%5C%5C%0A+%5Cfrac%7Btg%5E2x%2B1-2tg%5C+x%7D%7Btg%5C+x%7D%3D0%5C%5C%0Atg%5E2x-2tg%5C+x%2B1%3D0%5C%5C%0AD%3D4-4%3D0%5C%5C%0Atg%5C+x%3D+%5Cfrac%7B2%7D%7B2%7D%3D1%5C%5C%0Ax%3D+%5Cfrac%7B%5Cpi%7D%7B4%7D%2B%5Cpi+n%2C+n%5Cin+%5Cmathcal+Z++++)
На отрезке [-2π; π]:
-7π/4, -3π/4, π/4, то есть
три корня.
3. На фотке
4. На фотке (0 корней)
BL - высота, биссектриса, медиана.
S=1/2*BL*AC
AC=20√2
AL=1/2AC=10<span>√2
</span>В треугольнике BLC:
за теоремой Пифагора: BC=<span>√BL^2+LC^2
BC=</span><span>√225=15
Ответ : 15 </span>