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Ответ дал:
0
4.
ОДЗ: x > 1,5
![log_{1,5}(x-1) - log_{1,5}(2x-3) = 1\
log_{1,5} frac{x-1}{2x-3} =1\
frac{x-1}{2x-3}= frac{3}{2} \
6x - 9 = 2x-2\
4x = 7\
x = 1,75 log_{1,5}(x-1) - log_{1,5}(2x-3) = 1\
log_{1,5} frac{x-1}{2x-3} =1\
frac{x-1}{2x-3}= frac{3}{2} \
6x - 9 = 2x-2\
4x = 7\
x = 1,75](https://tex.z-dn.net/?f=log_%7B1%2C5%7D%28x-1%29+-+log_%7B1%2C5%7D%282x-3%29+%3D+1%5C%0Alog_%7B1%2C5%7D++frac%7Bx-1%7D%7B2x-3%7D+%3D1%5C%0A+frac%7Bx-1%7D%7B2x-3%7D%3D+frac%7B3%7D%7B2%7D+%5C%0A6x+-+9+%3D+2x-2%5C%0A4x+%3D+7%5C%0Ax+%3D+1%2C75)
5.
![2log_9^2x + 3log_9x-2=0\
log_9x=t\
2t^2+3t-2=0\
D=9+16=25\
t_1= frac{-3-5}{4} = -2\
t_2= frac{-3+5}{4} =0,5\
x_1=9^{-2} = frac{1}{81} \
x_2 = 9^{0,5}=3 2log_9^2x + 3log_9x-2=0\
log_9x=t\
2t^2+3t-2=0\
D=9+16=25\
t_1= frac{-3-5}{4} = -2\
t_2= frac{-3+5}{4} =0,5\
x_1=9^{-2} = frac{1}{81} \
x_2 = 9^{0,5}=3](https://tex.z-dn.net/?f=2log_9%5E2x+%2B+3log_9x-2%3D0%5C%0Alog_9x%3Dt%5C%0A2t%5E2%2B3t-2%3D0%5C%0AD%3D9%2B16%3D25%5C%0At_1%3D+frac%7B-3-5%7D%7B4%7D+%3D+-2%5C%0At_2%3D+frac%7B-3%2B5%7D%7B4%7D+%3D0%2C5%5C%0Ax_1%3D9%5E%7B-2%7D+%3D++frac%7B1%7D%7B81%7D+%5C%0Ax_2+%3D+9%5E%7B0%2C5%7D%3D3)
Ответ: 3
ОДЗ: x > 1,5
5.
Ответ: 3
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