Question

Найти область определения функции y=lg(4-5x)/(x-3)

Answer 1

$\frac{4-5x}{x-3} 0 \\ \\ x \neq 3 \\ \left \{ {{4-5x} 0\atop {x-30}} \right. \\ \left \{ {{5x<4} \atop {x3}} \right. \\ \left \{ {{x<0.8} \atop {x3}} \right.$

решений нет

$\frac{4-5x}{x-3} 0 \\ \\ x \neq 3 \\ \\ \left \{ {{4-5x<0} \atop {x-3<0}} \right. \\ \left \{ {{5x4} \atop {x<3}} \right. \\ \\ \left \{ {{x0.8} \atop {x<3}} \right.$

ответ:0.8<x<3.

Answer 2

$\\\frac{4-5x}{x-3}0/*(x-3)^2, \ x\neq3 \\ \\(4-5x)(x-3)0 \\ \\x_o=\frac45 \ \vee \ x_o=3 \\ \\x\in(\frac45,3)$