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

Question

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

Answer 1

0 \\ \\ x \neq 3 \\ \left \{ {{4-5x} >0\atop {x-3>0}} \right. \\ \left \{ {{5x<4} \atop {x>3}} \right. \\ \left \{ {{x<0.8} \atop {x>3}} \right. " alt=" \frac{4-5x}{x-3} >0 \\ \\ x \neq 3 \\ \left \{ {{4-5x} >0\atop {x-3>0}} \right. \\ \left \{ {{5x<4} \atop {x>3}} \right. \\ \left \{ {{x<0.8} \atop {x>3}} \right. " align="absmiddle" class="latex-formula">

решений нет

0 \\ \\ x \neq 3 \\ \\ \left \{ {{4-5x<0} \atop {x-3<0}} \right. \\ \left \{ {{5x>4} \atop {x<3}} \right. \\ \\ \left \{ {{x>0.8} \atop {x<3}} \right." alt=" \frac{4-5x}{x-3} >0 \\ \\ x \neq 3 \\ \\ \left \{ {{4-5x<0} \atop {x-3<0}} \right. \\ \left \{ {{5x>4} \atop {x<3}} \right. \\ \\ \left \{ {{x>0.8} \atop {x<3}} \right." align="absmiddle" class="latex-formula">

Ответ:0.8

Answer 2

0/*(x-3)^2, \ x\neq3 \\ \\(4-5x)(x-3)>0 \\ \\x_o=\frac45 \ \vee \ x_o=3 \\ \\x\in(\frac45,3)" alt="\\\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)" align="absmiddle" class="latex-formula">