Lg(2−3x)+lg(2+3x)=lg(4−x)+lgx

\left \{ {{4-x\ \textgreater \ 0} \atop {x\ \textgreater \ 0}} \right.{x > 04−x > 0

\left \{ {{x\ \textless \ 4} \atop {x\ \textgreater \ 0}} \right.{x > 0x < 4 <br>xx (0; \frac{2}{3} )(0;32)
lg[(2-3x)(2+3x)]=lg[(4-x)x]lg[(2−3x)(2+3x)]=lg[(4−x)x]
lg(4-9x^2)=lg(4x-x^2)lg(4−9x2)=lg(4x−x2)
4-9x^2=4x-x^24−9x2=4x−x2
8x^2+4x-4=08x2+4x−4=0
2x^2+x-1=02x2+x−1=0
D=1^2-4*2*(-1)=9D=12−4∗2∗(−1)=9
x_1= \frac{-1+3}{4}=0.5x1=4−1+3=0.5
x_2= \frac{-1-3}{4}=-1x2=4−1−3=−1 ∅
Ответ: 0.5