Question

Решить неравенство а)(х-2)log₀,₅х≤o
b) xˣ∠1 ,x больше 0
с) lg(x-2)+lg(27-x)∠2

Answer 1

$1); ; (x-2)log_{0,5}x leq 0; ,; ; ODZ:; ; x textgreater 0\\ left { {{x-2 leq 0} atop {log_{0,5}x geq 0}} right. ; ; ; ili; ; ; left { {{x-2 geq 0} atop {log_{0,5}x leq 0}} right. \\ left { {{x leq 2} atop {0 textless x leq 1}} right. ; ; ; ; ; ili; ; ; ; ; left { {{x geq 2} atop {x geq 1}} right. \\0 textless x leq 1; ; ; ; ili; ; ; ; ; x geq 2\\Otvet;; ; xin (0,1, ]cup [, 2,+infty )$

$2); ; x^{x} textless 1; ; ,; ; ; ; x textgreater 0\\xlgx textless lg1\\xlgx textless 0\\ left { {{x textgreater 0} atop {lgx textless 0}} right. ; left { {{x textgreater 0} atop {x textless 1}} right. \\Otvet:; ; xin (0,1).$

$3); ; lg(x-2)+lg(27-x) textless 2; ,; ; ; ODZ:; ; left { {{x-2 textgreater 0} atop {27-x textgreater 0}} right. ; left { {{x textgreater 2} atop {x textless 27}} right. ; to \\xin (2,27)\\lg(x-2)(27-x) textless lg10^2\\27x-x^2-54+2x textless 100\\x^2-29x+154 textgreater 0\\D=225; ,; ; x_1=frac{29-15}{2}=7,; ; x_2=22\\Otvet:; ; x_1=7,; x_2=22.$