AB (1; -2; 3)
AC = (3; 0; 1)
![|AB| = \sqrt{1^2+(-2)^2+3^2} = \sqrt{14} \\ |AC|= \sqrt{3^2+0^2+1^2}= \sqrt{10}](https://tex.z-dn.net/?f=%7CAB%7C+%3D+%5Csqrt%7B1%5E2%2B%28-2%29%5E2%2B3%5E2%7D+%3D+%5Csqrt%7B14%7D+%5C%5C+%7CAC%7C%3D+%5Csqrt%7B3%5E2%2B0%5E2%2B1%5E2%7D%3D+%5Csqrt%7B10%7D++)
AB·AC = 1·3 + (-2)·0 + 3·1 = 6
![cos \alpha = \frac{AB*AC}{|AB|*|AC|} = \frac{6}{ \sqrt{14}* \sqrt{10} } =\frac{3}{ \sqrt{35} }](https://tex.z-dn.net/?f=cos+%5Calpha+%3D+%5Cfrac%7BAB%2AAC%7D%7B%7CAB%7C%2A%7CAC%7C%7D+%3D+%5Cfrac%7B6%7D%7B+%5Csqrt%7B14%7D%2A+%5Csqrt%7B10%7D++%7D++%3D%5Cfrac%7B3%7D%7B+%5Csqrt%7B35%7D+%7D+)
4х²+36х+32=0
D=36²-4·4·32=1296-512=784=28²
x=(-36-28)/8=- 8 или х=(-38+28)/8=-5/4
Значит
![4 x^{2} +36x+32=4(x-(-8))\cdot (x-(- \frac{5}{4}))=(x+8)(5x+4), \\ \frac{x+8}{4 x^{2} +36x+32}= \frac{x+8}{(x+8)(5x+4)} = \frac{1}{5x+4}](https://tex.z-dn.net/?f=4+x%5E%7B2%7D+%2B36x%2B32%3D4%28x-%28-8%29%29%5Ccdot+%28x-%28-+%5Cfrac%7B5%7D%7B4%7D%29%29%3D%28x%2B8%29%285x%2B4%29%2C+%5C%5C++%5Cfrac%7Bx%2B8%7D%7B4+x%5E%7B2%7D+%2B36x%2B32%7D%3D+%5Cfrac%7Bx%2B8%7D%7B%28x%2B8%29%285x%2B4%29%7D+%3D+%5Cfrac%7B1%7D%7B5x%2B4%7D++)
Наименьшее на указанном промежутке равно 5 (при х=28)
1задание 3задание 4задание
1)-3 1)1
2)3/2 2)64/16=4 1)4-5х>0 5x<4 x<0.8
3)0
4)7 2)x-1>0 x>1
5)9 x-1неравно 1 х неравно 2
6)16*5=80 4-x>0 x<4 (1;2) u (2;4)