100 БАЛЛОВ!!! Есть идеи как это решать?
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Ответил IrkaShevko
0
ОДЗ: 1 - 2x > 0; 1 - 3x > 0; 1 - 2x ≠ 1; 1 - 3x ≠ 1 => x < 1/3; x≠0
![log_{1-2x}((1-2x)(1-3x))-log_{1-3x}(1-2x)^2 = 2\
log_{1-2x}(1-2x) + log_{1-2x}(1-3x)-2log_{1-3x}(1-2x)=2\
1 + log_{1-2x}(1-3x)- frac{2}{log_{1-2x}(1-3x)} =2\
log_{1-2x}(1-3x)=a\
a - frac{2}{a} =1\
a^2 - a - 2 = 0\
D=1 + 8 = 9\
a_1 = frac{1- 3 }{2}=-1 \
a_2= frac{1+3}{2} =2\
log_{1-2x}(1-3x)=-1\
frac{1}{1-2x} =1-3x\
6x^2-5x+1=1\
x(6x-5)=0\
x_1=0\
x_2= frac{5}{6} textgreater frac{1}{3} \
log_{1-2x}(1-3x)=2\
(1-2x)^2=(1-3x)\
4x^2-4x+1=1-3x\
4x^2-x=0\
x(4x-1)=0\
x_3=0\
x_4= frac{1}{4}](https://tex.z-dn.net/?f=log_%7B1-2x%7D%28%281-2x%29%281-3x%29%29-log_%7B1-3x%7D%281-2x%29%5E2+%3D+2%5C%0Alog_%7B1-2x%7D%281-2x%29+%2B+log_%7B1-2x%7D%281-3x%29-2log_%7B1-3x%7D%281-2x%29%3D2%5C%0A1+%2B+log_%7B1-2x%7D%281-3x%29-+frac%7B2%7D%7Blog_%7B1-2x%7D%281-3x%29%7D+%3D2%5C%0Alog_%7B1-2x%7D%281-3x%29%3Da%5C%0Aa+-++frac%7B2%7D%7Ba%7D+%3D1%5C%0Aa%5E2+-+a+-+2+%3D+0%5C%0AD%3D1+%2B+8+%3D+9%5C%0Aa_1+%3D++frac%7B1-+3+%7D%7B2%7D%3D-1+%5C%0Aa_2%3D+frac%7B1%2B3%7D%7B2%7D+%3D2%5C%0Alog_%7B1-2x%7D%281-3x%29%3D-1%5C%0A+frac%7B1%7D%7B1-2x%7D+%3D1-3x%5C%0A6x%5E2-5x%2B1%3D1%5C%0Ax%286x-5%29%3D0%5C%0Ax_1%3D0%5C%0Ax_2%3D+frac%7B5%7D%7B6%7D++textgreater+++frac%7B1%7D%7B3%7D+%5C%0Alog_%7B1-2x%7D%281-3x%29%3D2%5C%0A%281-2x%29%5E2%3D%281-3x%29%5C%0A4x%5E2-4x%2B1%3D1-3x%5C%0A4x%5E2-x%3D0%5C%0Ax%284x-1%29%3D0%5C%0Ax_3%3D0%5C%0Ax_4%3D+frac%7B1%7D%7B4%7D+ "log_{1-2x}((1-2x)(1-3x))-log_{1-3x}(1-2x)^2 = 2\
log_{1-2x}(1-2x) + log_{1-2x}(1-3x)-2log_{1-3x}(1-2x)=2\
1 + log_{1-2x}(1-3x)- frac{2}{log_{1-2x}(1-3x)} =2\
log_{1-2x}(1-3x)=a\
a - frac{2}{a} =1\
a^2 - a - 2 = 0\
D=1 + 8 = 9\
a_1 = frac{1- 3 }{2}=-1 \
a_2= frac{1+3}{2} =2\
log_{1-2x}(1-3x)=-1\
frac{1}{1-2x} =1-3x\
6x^2-5x+1=1\
x(6x-5)=0\
x_1=0\
x_2= frac{5}{6} textgreater frac{1}{3} \
log_{1-2x}(1-3x)=2\
(1-2x)^2=(1-3x)\
4x^2-4x+1=1-3x\
4x^2-x=0\
x(4x-1)=0\
x_3=0\
x_4= frac{1}{4}")
С учетом ОДЗ
Ответ: х = 0,25
С учетом ОДЗ
Ответ: х = 0,25
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