Уравнение прямой АС: (x-3)/(5-3)=(y-8)/(0-8)
(x-3)/2=(y-8)/(-8) |*2
x-3=(y-8)/(-4)
-4x+12=y-8
y=-4x+22
точка В(9;t)∈AC, ⇒
t=-4*9+22
t=-36+22
t=-14
ответ:при t=-14 точки А(3;8), С(5;0), В(9;-14) лежат на обной прямой
0,5х+6=8
0,5х=8-6
0,5х=2
х=2:0,5
х=4
3/11 = 0,(27) = 0,27 - округлено до сотых
![5*4 ^{x} + 3*10 ^{x} = 2* 25 ^{x}](https://tex.z-dn.net/?f=5%2A4+%5E%7Bx%7D+%2B+3%2A10+%5E%7Bx%7D+%3D+2%2A+25+%5E%7Bx%7D+)
Разделим обе части на
![25 ^{x}](https://tex.z-dn.net/?f=25++%5E%7Bx%7D++)
![5*( \frac{4}{25}) ^{x} + 3*( \frac{10}{25} ) ^{x} - 2*( \frac{25}{25}) ^{x}= 0](https://tex.z-dn.net/?f=5%2A%28+%5Cfrac%7B4%7D%7B25%7D%29+%5E%7Bx%7D+%2B+3%2A%28+%5Cfrac%7B10%7D%7B25%7D+%29+%5E%7Bx%7D+-+2%2A%28+%5Cfrac%7B25%7D%7B25%7D%29+%5E%7Bx%7D%3D+0+++)
![5*( \frac{2}{5}) ^{2x} +3*( \frac{2}{5}) ^{x} - 2 = 0](https://tex.z-dn.net/?f=5%2A%28+%5Cfrac%7B2%7D%7B5%7D%29+%5E%7B2x%7D+%2B3%2A%28+%5Cfrac%7B2%7D%7B5%7D%29+%5E%7Bx%7D+-+2+%3D+0++)
Обозначим
![( \frac{2}{5}) ^{x}= m\ \textgreater \ 0](https://tex.z-dn.net/?f=%28+%5Cfrac%7B2%7D%7B5%7D%29+%5E%7Bx%7D%3D+m%5C+%5Ctextgreater+%5C++0++)
![5 m^{2} + 3m - 2 = 0](https://tex.z-dn.net/?f=5+m%5E%7B2%7D+%2B+3m+-+2+%3D+0)
D = 3² - 4*5*(-2) = 9 + 40 = 49
m₁ = (- 3 + √49)/10 = (- 3 + 7)/10 = 2/5
m₂ = (- 3 - √49)/10 = (- 3 - 7)/10 = - 1 - посторонний корень
![( \frac{2}{5} ) ^{x}= \frac{2}{5}](https://tex.z-dn.net/?f=%28+%5Cfrac%7B2%7D%7B5%7D+%29+%5E%7Bx%7D%3D++%5Cfrac%7B2%7D%7B5%7D++)
x = 1
![\ln(x^3-7x+2\sin x+3)=\ln(x^3-7x+2\sin x-4)](https://tex.z-dn.net/?f=%5Cln%28x%5E3-7x%2B2%5Csin+x%2B3%29%3D%5Cln%28x%5E3-7x%2B2%5Csin+x-4%29)
Пусть
![x^3-7x+2\sin x=t](https://tex.z-dn.net/?f=x%5E3-7x%2B2%5Csin+x%3Dt)
, тогда получаем
![\ln (t+3)=\ln (t-4)\\ t+3=t-4\\ 0=-7](https://tex.z-dn.net/?f=%5Cln+%28t%2B3%29%3D%5Cln+%28t-4%29%5C%5C+t%2B3%3Dt-4%5C%5C+0%3D-7)
Откуда не тождество, а значит уравнение решений не имеет.
Ответ: нет решений.
![\log_2( \sqrt{x-1}+ \sqrt{1-x} +2)=\log_2^7x+1](https://tex.z-dn.net/?f=%5Clog_2%28+%5Csqrt%7Bx-1%7D%2B+%5Csqrt%7B1-x%7D++%2B2%29%3D%5Clog_2%5E7x%2B1)
ОДЗ:
![\begin{cases} & \text{ } 1-x \geq 0 \\ & \text{ } x-1 \geq 0 \\ & \text{ } \sqrt{1-x}+ \sqrt{1-x}+2 \ \textgreater \ 0 \\ & \text{ } 1-x \geq 0 \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%0A%26+%5Ctext%7B+%7D+1-x+%5Cgeq+0+%5C%5C+%0A%26+%5Ctext%7B+%7D+x-1+%5Cgeq+0+%5C%5C+%0A%26+%5Ctext%7B+%7D++%5Csqrt%7B1-x%7D%2B+%5Csqrt%7B1-x%7D%2B2+%5C+%5Ctextgreater+%5C+0+++%5C%5C+%0A%26+%5Ctext%7B+%7D+1-x+%5Cgeq+0+%0A%5Cend%7Bcases%7D)
так как
![\begin{cases} & \text{ } x-1 \geq 0 \\ & \text{ } 1-x \leq 0 \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%0A%26+%5Ctext%7B+%7D+x-1+%5Cgeq+0+%5C%5C+%0A%26+%5Ctext%7B+%7D+1-x+%5Cleq+0+%0A%5Cend%7Bcases%7D)
, то можно сделать уравнение таким образом
![\begin{cases} & \text{ } x\ \textgreater \ 0 \\ & \text{ } 1-x=0 \\ & \text{ } \sqrt{x-1}+ \sqrt{1-x}+2\ \textgreater \ 0 \\ & \text{ } \log_2( \sqrt{x-1}+ \sqrt{1-x}+2)=\log_2^7x+1 \end{cases}\Rightarrow\begin{cases} & \text{ } 1\ \textgreater \ 0 \\ & \text{ } x=1 \\ & \text{ } 2\ \textgreater \ 0 \\ & \text{ } 1=1 \end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%0A%26+%5Ctext%7B+%7D+x%5C+%5Ctextgreater+%5C+0+%5C%5C+%0A%26+%5Ctext%7B+%7D+1-x%3D0+%5C%5C+%0A%26+%5Ctext%7B+%7D++%5Csqrt%7Bx-1%7D%2B+%5Csqrt%7B1-x%7D%2B2%5C+%5Ctextgreater+%5C+0+++%5C%5C+%0A%26+%5Ctext%7B+%7D+%5Clog_2%28+%5Csqrt%7Bx-1%7D%2B+%5Csqrt%7B1-x%7D%2B2%29%3D%5Clog_2%5E7x%2B1+++%0A%5Cend%7Bcases%7D%5CRightarrow%5Cbegin%7Bcases%7D%0A%26+%5Ctext%7B+%7D+1%5C+%5Ctextgreater+%5C+0+%5C%5C+%0A%26+%5Ctext%7B+%7D+x%3D1+%5C%5C+%0A%26+%5Ctext%7B+%7D+2%5C+%5Ctextgreater+%5C+0+%5C%5C+%0A%26+%5Ctext%7B+%7D+1%3D1+%0A%5Cend%7Bcases%7D)
Ответ: x=1