Question

Здравствуйте, помогите решить

Answer 1

Ответ:

Объяснение:

$\displaystyle\\\left \{ {{\dfrac{2x-1}{4}+3-5x<4 } \atop {\dfrac{x^2-7x}{1-x}\leq0 }} \right.\\\\\\\left \{ {{\dfrac{2x-1}{4}-5x-1<0 } \atop {\dfrac{x(x-7)}{x-1}\geq 0 }}\right.\\\\\\ \ \left \{ {2x-1-20x-4<0 } \atop {\dfrac{x(x-7)}{x-1}\geq 0 }} \right.\\\\\\\left \{ {18x>-5} \atop {\dfrac{x(x-7)}{x-1}\geq 0 }} \right.\ \ \ \ \ \ \ ---[0]+++(1)---[7]+++\\\\\\\left \{ {{x>-5/18} \atop {x\in[0;1)U[7;+\infty)}} \right. \\\\\\Otvet:x\in[0;1)U[7;+\infty)}$