Question

log2 (х-1) + log2 Х < 1 Решите неравество

Answer 1

$\log_{2}(x-1)+\log_2{x} \ \textless \ 1 \\ x(x-1)\ \textless \ 2 \\ x^2-x-2 \ \textless \ 0 \\ x_1=-1, x_2=2 \\ x \in (-1, 2)$

Answer 2

$log_{2} x(x-1)\ \textless \ 1$
$log_{2} x^{2} -x\ \textless \ 1$
$log_{2} x^{2} -x\ \textless \ log_{2} 2$
x²-x<2
x²-x-2<0
x²-x-2=0
по теореме Виета
x₁=2
x₂=-1

рисуем луч
x∈(-1;2)