Question

lg(2-3x)+lg(2+3x)=lg(4-x)+lg(x)

Answer 1

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​ 
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