Foreword
Related concepts
The form \ (2 <2x + 1 < 3 \) inequality, we call double inequality.
A method for solving a double inequality, using the properties of solving inequalities, double inequality to the left, center, right Subtracting \ (1 \) , to give \ (1 <2x <2 \) , then simultaneously divide \ (2 \) , to give \ (\ cfrac {. 1} {2} <X <. 1 \) ; second method, into Inequalities Solution, such as \ (\ left \ {\ begin {array} {l} {2 < 2x + 1} \\ {2x + 1 <3.} \ end {array} \ right. \)
Typical Example Analysis
Method 1: Inequality equivalent to the original \ (\ left \ {\ begin {array} {l} {0 <\ cfrac {1 + lga} {1-lga} ①} \\ {\ cfrac {1 + lga} { 1-lga} <1②} \ end {array} \ right. \)
Solutions ① \ (0 <\ cfrac {. 1 + LGA} {. 1-LGA} \) , obtained from the worn roots method \ (\ cfrac {. 1 + LGA} {LGA-. 1} <0 \) , so \ (--1 <LGA <. 1 \) ③,
Solutions ② \ (\ cfrac {. 1 + LGA} {. 1-LGA} <. 1 \) , obtained by deforming \ (\ cfrac {2lga} {LGA-. 1}> 0 \) , obtained from the worn roots method \ (lga <0 \) or \ (LGA> 1 \) ④,
Therefore, seeking to obtain from the intersection ③④ \ (-. 1 <LGA <0 \) , solve for \ (A \ in (\ cfrac {} {10}. 1,. 1) \) .
Method 2: see the middle of the fractional portion of the double inequality, if used to associate a fractional modification, this may be solved;
By the \ (0 <\ cfrac {. 1 + LGA} {. 1-LGA} <. 1 \) , to give \ (0 <\ cfrac {LGA-. 1 + 2} {. 1-LGA} <. 1 \) , i.e. \ (0 <-1 + \ {2} {cfrac-LGA. 1} <. 1 \) , so \ (. 1 <\ {2} {cfrac-LGA. 1} <2 \) , and can give \ (1-lga> 0 \ ) ,
Therefore, using the reciprocal law give \ (\ cfrac {. 1} {2} <\ cfrac {. 1-LGA} {2} <. 1 \) , i.e. \ (. 1 <. 1-LGA <2 \) , i.e. \ (- 2 < . 1-LGA <-1 \) , i.e. \ (-. 1 <LGA <0 \) , to give solution to give solution \ (A \ in (\ cfrac {} {10}. 1,. 1) \) , was chosen \ (C \) .