1958 AMC 12 第 34 题

Question

一个分数的分子为

6x+1

，分母为

7-4x

，且

x

可取从

-2

到

2

的任意值，包括两端。使分子大于分母的

x

值为：

The numerator of a fraction is

6x+1,

the denominator is

7-4x,

and

x

can have any value between

-2

and

2,

both included. The values of

x

for which the numerator is greater than the denominator are:

\dfrac35\lt x\le2

\dfrac35\le x\le2

0\lt x\le2

0\le x\le2

-2\le x\le2

直接解不等式

6x+1\gt7-4x

来比较分子与分母

Compare the numerator and denominator directly by solving

6x+1\gt7-4x

将所得不等式的解集与给定区间

[-2,2]

取交集

Intersect the resulting inequality with the given interval

[-2,2]

所需比较给出

6x+1\gt7-4x\text{，}

所以

10x\gt6

，且

x\gt\frac{3}{5}

。与

-2\le x\le2

取交集得到

\frac35\lt x\le2\text{。}

所以正确答案为 A。

The required comparison gives

6x+1\gt7-4x,

so

10x\gt6

and

x\gt\frac{3}{5}.

Intersecting this with

-2\le x\le2

gives

\frac35\lt x\le2.

Thus, the correct answer is A.

其他年份的第 34 题