ОДЗ
{9-x≥0⇒x≤9
{(x-3)³>0⇒x-3>0⇒x>3
{x-3≠1⇒x≠4
x∈(3;4) U (4;9]
1)x∈(3;4) основание меньше 1,знак меняется
9-x≤1
x≥8
нет решения
2)x∈(4;9]
x≤8
Ответ x∈(4;8]
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+++++++++++++++)
D=b^2-4ac=4+12=16
y=-b±√D/2a=(-2±4)/2
первый корень=1
второй корень=-3
Ответ:
а) х = 1/4 = 0,25; б) х = 4; в) х1 = -3; х2 = 8
Объяснение:
а) 1/х + 5х/(х + 1) = 5
ОДЗ: х ≠0; х ≠ -1
х + 1 + 5х² = 5х² + 5х
4х = 1
х = 1/4
б) (3х² - 48)/(х + 4) = 0
ОДЗ: х ≠ -4
3х² - 48 = 0
х² = 16
х1 = -4 (не подходит из-за ОДЗ
х2 = 4
в) 10/(х - 3) - 8/х = 1
ОДЗ: х ≠ 3; х ≠ 0
10х - 8х + 24 = х² - 3х
х² - 5х - 24 = 0
D = 5² - 4· (-24) = 121
√D= 11
х1 = (5 - 11)/2 = -3
х2 = (5 + 11)/2 = 8
1) a_(n+1)=2((n+1)-10)=2n-18
a_n+2=2(n+2-10)=2n-16
a_n+5=2(n+5-10)=2n-10
2) a_n=7*2^(-n-2)
a_n+1=7*2^(-n-1-2)=7*2^(-n-3)
a_n+2=7*2^(-n-4)
a_n+5=7*2^(-n-7)