Win the ninth grade mathematics at the starting line of the senior high school entrance examination

fill-in-the-blank question

1.(20 12 Panzhihua, Sichuan, 4 points) If the fractional equation has roots, then k = ▲.

Answer 1.

Rooting of fractional equation in test center.

This analysis is obtained by solving the fractional equation.

The fractional equation has increasing roots,

The solution of 2-x = 0-x-2 = 0 is: x=2. The solution is: k= 1.

2.(20 12 Yibin, Sichuan, 3 points) The solution of the linear inequality group is ▲.

Answer -3 ≤ x

Try to solve a set of linear inequalities.

Analyze and solve a linear inequality group, first find the solution set of each inequality in the inequality group, and then use formulas to find the formulaic part of these solution sets: take the big one, take the small one, find the middle between the big one and the small one, and the big one and the small one cannot be solved (no solution). Therefore,

From the first inequality x≥﹣3, from the second inequality x < ﹣ 1, the solution set of the inequality group is ﹣ 3 ≤ x < ﹣ 1.

3. The positive integer solution of inequality 2x+9≥3(x+2) is ▲.

Answer 1, 2,3.

Integer solution of one-dimensional linear inequality in test sites.

Analyze the inequality first, find its solution set, and then judge its positive integer solution according to the solution set:

2x+9≥3(x+2), brackets removed, 2x+9≥3x+6, transposed, 2x-3x ≥ 6-9, combined with similar terms, x ≥ 3,

If the coefficient is 1, x≤3. ∴ Its positive integer solution is 1, 2,3.

4.(20 12, Dazhou, Sichuan, 3 points) If the solution of the binary linear equations of x and y satisfies X+Y > 1, then the value range of k is ▲.

The answer k > 2.

Test points to solve binary linear equations and univariate linear inequalities.

Analyze and solve the equations about X and Y, use K to represent the values of X and Y, then substitute the values of X and Y into x+y > 1 to get the inequality about K, and find out the value range of K:

Solve.

∵ x+y > 1, ∴ 2k-k- 1 > 1, and the solution is k > 2.

5. (Mianyang, Sichuan, 20 12, 4 points) If the length of a rectangle is reduced by 5cm and the width is increased by 2cm to become a square, and the areas of these two figures are equal, then the area of the original rectangle is ▲ cm2.

The answer.

The application of one-dimensional linear equation in examination center (geometry problem).

Let the side length of a square be xcm, then (x+5) (x-2) = x2, and x= is obtained. ,

∴S= .

6. (Mianyang, Sichuan, 20 12, 4 points) If the integer solution of the inequality group x: has only 1 2, then there are ▲ ordered number pairs (a, b)*** suitable for this inequality group.

Answer 6.

Integer solution of one-dimensional linear inequality in test sites

Analysis,

From ①:; From 2:.

∵ Inequality group has a solution, and the solution set of∴ Inequality group is:

The integer solution of the inequality group is only 1, 2, as shown in the figure:

,

∴ 0 < ≤ 1, 2 ≤ < 3, the solution is: 0 < A ≤ 3, 4 ≤ B < 6.

∴a= 1,2,3,b=4,5。

∴ There are 3×2=6 ordered number pairs (a, b)*** consisting of integers A and B.

7.(20 12 Liangshan, Sichuan, 4 points) The price of a commodity is 528 yuan, and the profit that merchants can get from selling such a commodity is 10% ~ 20% of the purchase price. If the purchase price is X yuan, the value range of X is ▲.

The answer is 440≤x≤480.

The application of one-dimensional linear inequality system in examination center.

Analysis basis: price = purchase price ×( 1+ profit rate), we can get: purchase price = price 1+ profit rate, goods can make a profit (10% ~ 20%), that is, the price is at least (1+/kloc-0).

528 1+20%≤x≤528 1+ 10%，

The solution is 440≤x≤480.

The value range of ∴x is 440≤x≤480.

8.(20 12, Bazhong, Sichuan, 3 points) If the equation about x has roots, the value of m is ▲.

Answer 0.

Rooting of fractional equation in test center.

Both sides of the analytic equation are multiplied by the simplest common denominator (x-2), and the fractional equation is transformed into an integral equation, and then the root of the fractional equation is to make

The simplest common denominator is equal to 0. Find the value of x, and then substitute it into the calculation to find the value of m:

Both sides of the equation are multiplied by (x-2), and 2-x-m = 2 (x-2).

The fractional equation has an increasing root, ∴ x-2 = 0, and the solution is x=2.

∴ 2-2-m = 2 (2-2), and the solution is m=0.

9.(20 12, Ziyang, Sichuan, 3 points) If the unary quadratic equation about X has two unequal real roots, the range of k is ▲.

The answer is k 0. Then the equation about k is listed, and the equation can be solved by combining the definition of quadratic equation in one variable:

∫ There are two unequal real roots,

∴△ = 1-4k > 0, and k≠0, the solution is that the solution set of k 6 is ▲

The answer x > 4.

Try to solve the one-dimensional inequality.

By moving the inequality from left to right, it can be directly combined and solved: moving the term from x>6-2 > 6: x+2>6-2, the solution can be obtained: x > 4.