Question

С-26 ВАРИАНТ 2
Решите плиз

Answer 1

$1.\a);log_5(3x-4)=log_5(12-5x)\O.D.3.:\begin{cases}3x-4 textgreater 0\12-5x textgreater 0end{cases}Rightarrow xinleft(1frac13;;2frac25right)\3x-4=12-5x\3x+5x=12-4\8x=8\x=1;-;He;nogx.;no;O.D.3.\KopHeu;HET.\b);log_3(x^2+3x-7)=1\x^2+3x-7=3^1\x^2+3x-10=0\D=9+4cdot10=49\x_{1,2}=frac{-3pm7}2\x_1=-5,;x_2=2$
$c);lg(x-1)+lg(x+1)=lg(9x+9)\begin{cases}x-1 textgreater 0\x+1 textgreater 0\9x+9 textgreater 0end{cases}Rightarrow x textgreater 1\lg(x-1)(x+1)=lg(9x+9)\x^2-1=9x+9\x^2-9x-10=0\D=81+4cdot10=121\x_{1,2}=frac{9pm11}2\x_1=10,\x_2=-1;-;He;nogx.;no;O.D.3.\\2.;lg(3x+10)=lg(3(4x-3)+10)\3x+10=12x+1\12x-3x=10-1\9x=9\x=1$