Y=5x-3
A) (1;8) не принадлежит
5•1-3=8
2=3
Б) (2;7) принадлежит
5•2-3=7
7=7
В) (-2;-7) не принадлежит
5•(-2)-3=-7
-13=-7
Г) (-1;2) не принадлежит
5•(-1)-3=2
-8=2
Решение
I 3x + 6 I = 12
3x + 6 = - 12 или 3x + 6 = 12
3x = - 12 - 6 3x = 12 - 6
3x = - 18 3x = 6
x1 = - 6 x2 = 2
По условию 3π/2<α<2π
это IV четверть где cosα>0 и sinα<0
значит cosα=¹²/₁₃
найти tg(π/4-α)
решение:
![\displaystyle sin^2a+cos^2a=1\\\\sina= \sqrt{1-cos^2a}= \sqrt{1-( \frac{12}{13})^2}= \sqrt{1- \frac{144}{169}}= \sqrt{ \frac{25}{169}}=| \frac{5}{13}|\\\\sina\ \textless \ 0\\\\sina=- \frac{5}{13}\\\\tga= \frac{sina}{cosa}= \frac{-5}{13}: \frac{12}{13}=- \frac{5}{12}\\\\tg( \frac{ \pi }{4}-a)= \frac{tg( \pi /4)-tga}{1+tg( \pi /4)*tga}= \frac{1-( \frac{-5}{12})}{1+1( \frac{-5}{12})}=\\\\= \frac{ \frac{17}{12}}{ \frac{7}{12}}= \frac{17}{7}](https://tex.z-dn.net/?f=%5Cdisplaystyle+sin%5E2a%2Bcos%5E2a%3D1%5C%5C%5C%5Csina%3D+%5Csqrt%7B1-cos%5E2a%7D%3D+%5Csqrt%7B1-%28+%5Cfrac%7B12%7D%7B13%7D%29%5E2%7D%3D+%5Csqrt%7B1-+%5Cfrac%7B144%7D%7B169%7D%7D%3D+%5Csqrt%7B+%5Cfrac%7B25%7D%7B169%7D%7D%3D%7C+%5Cfrac%7B5%7D%7B13%7D%7C%5C%5C%5C%5Csina%5C+%5Ctextless+%5C+0%5C%5C%5C%5Csina%3D-+%5Cfrac%7B5%7D%7B13%7D%5C%5C%5C%5Ctga%3D+%5Cfrac%7Bsina%7D%7Bcosa%7D%3D+%5Cfrac%7B-5%7D%7B13%7D%3A+%5Cfrac%7B12%7D%7B13%7D%3D-+%5Cfrac%7B5%7D%7B12%7D%5C%5C%5C%5Ctg%28+%5Cfrac%7B+%5Cpi+%7D%7B4%7D-a%29%3D+%5Cfrac%7Btg%28+%5Cpi+%2F4%29-tga%7D%7B1%2Btg%28+%5Cpi+%2F4%29%2Atga%7D%3D+%5Cfrac%7B1-%28+%5Cfrac%7B-5%7D%7B12%7D%29%7D%7B1%2B1%28+%5Cfrac%7B-5%7D%7B12%7D%29%7D%3D%5C%5C%5C%5C%3D+%5Cfrac%7B+%5Cfrac%7B17%7D%7B12%7D%7D%7B+%5Cfrac%7B7%7D%7B12%7D%7D%3D+%5Cfrac%7B17%7D%7B7%7D+++++++++++++++)
X²-xy-4x+4y=(x²-xy)-(4x-4y)=x(x-y)-4(x-y)=(x-y)(x-4)
ab-ac-bx+cx+c-b=(ab-ac)-(bx-cx)+(c-b)=a(b-c)-x(b-c)-(b-c)=(b-c)(a-x-1)
<span>-2ab³ * 3a² * b</span>⁴=-6a³b⁷
Bn=b1*q^(n-1)
162=2/9*3^(n-1)
3^(n-1)=81*9=3^4*3^2
n-1=6
n=7