5x-3=0
5x=3
x=3/5
))))))))))))))))))
Ваша задача решена ответ можете посмотрет в вложение
3/4 х =35- х х + 3/4 х = 35 , х = 35 :7/4 =20
х =20
![\log_{ \frac{1}{3} }(2+x)+\log_{ \frac{1}{3} }(5+4x)=0](https://tex.z-dn.net/?f=%5Clog_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D%282%2Bx%29%2B%5Clog_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D%285%2B4x%29%3D0)
Отметим ОДЗ:
![\left \{ {{2+x>0} \atop {5+4x>0}} \right. \to \left \{ {{x> \frac{-5}{4} } \atop {x>-2}} \right. \to x \in (- \frac{5}{4} ;+\infty)](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B2%2Bx%3E0%7D+%5Catop+%7B5%2B4x%3E0%7D%7D+%5Cright.+%5Cto+%5Cleft+%5C%7B+%7B%7Bx%3E+%5Cfrac%7B-5%7D%7B4%7D+%7D+%5Catop+%7Bx%3E-2%7D%7D+%5Cright.+%5Cto+x+%5Cin+%28-+%5Cfrac%7B5%7D%7B4%7D+%3B%2B%5Cinfty%29)
Воспользуемся свойством логарифмов
![\log_{ \frac{1}{3} }(2+x)+\log_{ \frac{1}{3} }(5+4x)=\log_{ \frac{1}{3} }1 \\ (2+x)(5+4x)=1 \\ 4x^2+13x+9=0](https://tex.z-dn.net/?f=%5Clog_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D%282%2Bx%29%2B%5Clog_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D%285%2B4x%29%3D%5Clog_%7B+%5Cfrac%7B1%7D%7B3%7D+%7D1+%5C%5C+%282%2Bx%29%285%2B4x%29%3D1+%5C%5C+4x%5E2%2B13x%2B9%3D0)
Находим дискриминант
![D=b^2-4ac=13^2-4\cdot4\cdot9=25](https://tex.z-dn.net/?f=D%3Db%5E2-4ac%3D13%5E2-4%5Ccdot4%5Ccdot9%3D25)
Воспользуемся формулой корней квадратного уравнения
![x_1_,_2= \dfrac{-b\pm \sqrt{D} }{2a}](https://tex.z-dn.net/?f=x_1_%2C_2%3D+%5Cdfrac%7B-b%5Cpm+%5Csqrt%7BD%7D+%7D%7B2a%7D+)
![x_1=- \frac{9}{4}](https://tex.z-dn.net/?f=x_1%3D-+%5Cfrac%7B9%7D%7B4%7D+)
- не удовлетворяет ОДЗ
![x_2=-1](https://tex.z-dn.net/?f=x_2%3D-1)
Ответ: -1.
![1+\log_5(x^2+4x-5)=\log_5(x+5)](https://tex.z-dn.net/?f=1%2B%5Clog_5%28x%5E2%2B4x-5%29%3D%5Clog_5%28x%2B5%29)
ОДЗ:
![\left \{ {{x+5>0} \atop {x^2+4x-5>0}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%2B5%3E0%7D+%5Catop+%7Bx%5E2%2B4x-5%3E0%7D%7D+%5Cright.+)
Воспользуемся свойством логарифмов
![\log_55+\log_5(x^2+4x-5)=\log_5(x+5) \\ 5(x^2+4x-5)=x+5 \\ 5x^2+19x-30=0](https://tex.z-dn.net/?f=%5Clog_55%2B%5Clog_5%28x%5E2%2B4x-5%29%3D%5Clog_5%28x%2B5%29+%5C%5C+5%28x%5E2%2B4x-5%29%3Dx%2B5+%5C%5C+5x%5E2%2B19x-30%3D0)
Опять же квадратное уравнение
![D=b^2-4ac=19^2+4\cdot5\cdot30=961](https://tex.z-dn.net/?f=D%3Db%5E2-4ac%3D19%5E2%2B4%5Ccdot5%5Ccdot30%3D961)
![x_1=-5](https://tex.z-dn.net/?f=x_1%3D-5)
- не удовлетворяет ОДЗ
![x_2= \frac{6}{5}](https://tex.z-dn.net/?f=x_2%3D+%5Cfrac%7B6%7D%7B5%7D+)
Ответ:
![\frac{6}{5}](https://tex.z-dn.net/?f=+%5Cfrac%7B6%7D%7B5%7D+)