The solution of high school mathematics score inequality is as follows: problem-solving idea: solve two unequal numbers on the left and right respectively, and then take the intersection of the two values. Preventive measures (error-prone places):

(1)x is preceded by a minus sign. When the negative sign moves to the other side of the inequality, the direction of the inequality will change (that is, the greater than sign becomes the less than sign, or the less than sign becomes the greater than sign).

(2) Because the numerator "2" is a positive number, if the score is greater than 0, it is only necessary to make the denominator greater than 0. To make the score less than 1, as long as the numerator of the score is greater than the denominator.

The denominator of shilling is not equal to zero, and then the main idea is to turn fractional inequality into algebraic expression inequality. When you look at the algebraic expression and the fraction together, you must first divide, and move 1 to the left of the inequality to get (x-1)/(2x+1)-(2x+1)/(2x+1).

Then continue the operation, (-x-2)/(2x+ 1) < =0, and it is still a fraction, which transforms two formulas of algebraic expression, (-x-2) * (2x+ 1).

Note that there are two formulas here, and we must pay attention to inequalities. If the original inequality has an equal sign, the algebraic expression score must not be equal to 0. If the original inequality has no equal sign, we don't need to consider these.

Rank the scores as (AX B)/(CX D) > 0 or (AX B)/(CX D) < 0. The former is (the same sign is positive), that is, solving the inequality group of AX B and CX D > 0 or < 0 at the same time (the first intersection is the union) and the latter is (the different sign is positive), that is.