Question

Решите неравенства
1. √(5-2x)<6x-1

2. log 3-2x x^2<=1

3.3√x-√(5x+5)>1

Answer 1

$1); ; sqrt{5-2x} textless 6x-1; ,; ; ; ODZ:5-2x geq 0; ,; x leq 2,5\\5-2x textless 36x^2-12x+1\\36x^2-10x-4 textgreater 0\\18x^2-5x-1 textgreater 0\\D=169; ,; ; x_1=-frac{2}{9}; ,; ; x_2= frac{1}{2} \\+++(-frac{2}{9})---(frac{1}{2})+++quad xin (-infty ,- frac{2}{9} )cup ( frac{1}{2},+infty )\\x leq 2,5; ; ; Rightarrow ; ; ; underline {xin (-infty ,-frac{2}{9})cup( frac{1}{2};; 2,5] }$

$2); ; log_{3-2x}x^2leq 1; ,; ; ; ODZ:; ; left { {{3-2x textgreater 0} atop {3-2xne 1}} right. ; ,left { {{x textless 1,5} atop {xne 1}} right.$

Метод рационализации:  $log_{h}f<1; Leftrightarrow ; (h-1)(f-h)<0$

$(3-2x-1)(x^2-3+2x) leq 0\\-2(x-1)(x-1)(x+3) leq 0; |cdot (-1/2)\\(x-1)^2(x+3) geq 0; ; ; ---[-3; ]+++(1)+++(1,5)\\underline {xin [-3;1)cdot (1;; 1,5)}$

$3)quad 3sqrt{x}-sqrt{5x+5} textgreater 1; ,; ; ; ODZ:; x geq 0\\3sqrt{x} textgreater 1+sqrt{5x+5}\\9x textgreater 1+2sqrt{5x+5}+5x+5\\2sqrt{5x+5} textless 4x-6; |:2\\sqrt{5x+5} textless 2x-3\\5x+5 textless 4x^2-12x+9\\4x^2-17x+4 textgreater 0\\D=225; ,; ; x_1=frac{1}{4}; ,; ; x_2=4\\quad [; 0; ]+++(frac{1}{4})---(4)+++\\underline {xin [; 0,frac{1}{4})cup (4,+infty )}$