A high school math problem about collections

Let the two roots be respectively, then x1+x2=-1 So at least one of the two roots is negative, if one element is a non-negative number, then x1 times x2 is equal to a less than or equal to 0, which is the solution to this problem.

At this point, the original equation always has a solution).

First of all, it cannot be an empty set, so the discriminant formula is greater than 0, 1-4a> or =0, and we get a" or =1 4

A=1 4, x=-1 2, so a<1 4

A complement where at least one element is a non-negative real number is that both elements are negative, and the sum of the two roots is -1, so the product of the two roots is greater than 0, i.e., a>0

So at least one element is a non-negative real number a" or = 0

In summary, a《or=0

Because at least one element is a non-negative real number.

So 0 1-4a 0 1 4 a

A complement where at least one element is a non-negative real number is that both elements are negative, and the sum of the two roots is -1, so the product of the two roots is greater than 0, i.e., a>0

So at least one element is a non-negative real number a" or = 0

In summary, a《or=0

x^2+x+a=0

At least one element is a non-negative real number.

Let the two be x1 x2

From the relationship between the root and the coefficient, x1+x2=-1 x1*x2=a, so from x1+x2=-1, we can know that at least one is negative.

So a is a real number.

However, the discriminant test should be greater than 0

So 1-4a>0

a<

It cannot be an empty set, the discriminant formula is greater than 0, and 1-4a> or =0 is obtained, and a" or =1 4 When a=1 4, x=-1 2, it can be concluded that a<1 4 has at least one element that is a non-negative real number complement is that all elements are negative, and the sum of the two roots is -1, so the product of the two roots is greater than 0, i.e., a>0

Therefore, at least one element is non-negative, and the real number is a" or = 0

So a《or=0

How else can you solve 1 in the classification discussion

As long as the equation x 2 + x + a = 0

f(0)<=0, discriminant 1 4a>=0

a<=0,a<=1/4

In summary, a<=0

According to the heterogeneity of the elements of the set.

x≠2-xx≠1

The set of values of x is descriptively expressed as ().

The mutuality of the elements x≠2-x introduces x≠1

x|x≠1}

There are two major cases:

1。Neither number is a multiple of 6.

You might as well make the AI an odd multiple of 3, 3, 9, 15, ,......99 has a total of 17 and has a chance of 17, while aj must be even, not a multiple of 6, 2, 4, 8, ......There are 34 possibilities in 100, 17*34=578.

2。One of the two numbers is a multiple of six.

Consider making the AI a multiple of six, 6, 12, 18, ......There are 16 possibilities for 96 and 100-16 = 84 for AJ.

16*84=1344 species.

3。Both numbers are multiples of six, then there are 16*15=240 kinds of numbers, and the sum of the above three numbers is the final result.

Solution: a===

Because a b=

When m+1 2m-1, i.e., b= time, m2 when m+1 2m-1, i.e., m2, 2m-1 -2 or m+1 5 solution gives m -1 2 or m 4

And because m 2, m 4

In summary, the value range of m is m 2 or m 4

b: x -5x+6 = 0 (x-2)(x-3) = 0 x = 2 or x = 3c: x +2x-8 = 0 (x-2) (x+4) = 0 x = 2 or x = -4

a∩b≠∅，a∩c=∅

x=3 satisfies a

Bring x=3 into a, 9-3a+a -19=0 a -3a-10=0 (a+2)(a-5)=0 to get a=-2 or a=5

After testing, a = -2

After solving a, bring a into the equation of a, see if it is the same as b and c, and if the formula is the same, then round off the value of a.

b==｛x|x=2 or x=3}

c==｛x|x=2 or x=-4}

a∩b≠∅，a∩c=∅

3 belongs to a and 2 does not belong to a

Substituting x=3 into a, 9-3a+a-19=0, i.e., a=-2 or a=5 substituting x=2 into a, 4-2a+a-19=0, i.e., a=-3 or a=5 a=-2

The equation must be solved, so a -4 (a -19) > = 0, that is, 3a < = 76 is tested to meet the conditions when a = -2.

a=-2