A math problem, ask the math emperor to come, all the math emperor ask for a math problem Thank you

To make proposition p or q a false proposition.

Then both propositions p and q are false propositions.

Proposition p: Equation a 2x 2 + ax-2 = 0 has a solution from -1 to 1.

The solution is: a 2x 2+ax-2=0

ax+2）(ax-1)=0

Solution: x=-2 a or x=1 a

To satisfy, there is a solution from -1 to 1, and to satisfy -2 a and 1 a, at least one value between -1 and 1 is satisfied.

Then there is: solution: -1<=-2 a<=1 , 1<=1 a<=1 (note: a is the denominator, so it cannot be 0).

Solution: When a>0, a>=2

When a <0, a<=-2

Solution: When a>0, a>=1

When a < 0, a<=-1

To sum up: the range of a is (- 1] [1,+

That is, in the above range, the proposition p is a true proposition, and conversely, in order to satisfy the proposition to make it a false proposition, the value range of a is 1 proposition q: there is only one real number x satisfying the inequality x 2+2ax+2a<=0

The solution is: x 2+2ax+2a<=0

x^2+2ax+a^2-a^2+2a<=0

x+a) ^2-(a^2-2a+1)+1<=0

x+a) ^2<=(a-1) ^2-1

Because the square of any number must be greater than or equal to 0, (x+a)2 must be greater than or equal to 0.

If the value of x is only one, then (x+a) 2=0 satisfies the meaning of the question, and it can be seen that when (a-1) 2-1=0, the proposition q is a true proposition, and the solution is: a=0 or a=2

That is, the mission question q is a false proposition, a satisfies not 0 and 2, and the synthesis can obtain: 1

Solution: x2-ax-2a2=0

x-2a)(x+a)=0

x=2a or x=-a

Because -1 x 2

So -1 2a 2 or -1 -a 2

-1 2 a 2 or -2 a 1

In summary, the value range of a is: -2 a 1

Since it is 21-2n under the root number, 21-2n is greater than or equal to 0, so n is an integer less than or equal to 21 2.

Since it is 7n-26 under the root number, 7n-26 is greater than or equal to 0, so n is an integer greater than or equal to 26 7.

So the manuscript is late: 26 7 n 21 2, n should be 4, 5, 6, 7, key empty Li 8, 9, 10

Satisfying the root number under 21-2n is an integer only 6,10

Satisfying the root number, 7n-26 is an integer, only the deficit is 5, and 6 is n=6 on the whole

This one is simple.

The sum of the two root numbers is an integer, so each of them is an integer.

The number under the root number is greater than or equal to 0, that is, 21-2n is greater than or equal to 0 and 7n-26 is greater than or equal to 0, then one by one, this brother Lu try, n makes the number under the root number both greater than or equal to 0 or the number that can be squared.

n can only be 6

21-2n>=0 and 7n-26>=0

26 7 < = n< = 21 2

Add n to be a natural number, then n is possible

Bring the n possible into the topic of the school of the Fu talk to give the hall lack of the style of the dust bumper. When n=6, it is an integer, so the natural number n=6 is sought

This is divided into situations, and I will briefly list it for you:

If you take 1 day off for 3 days, then the cycle is 4 days; That is, the number of days divided by the period can calculate the number of days off.

There are 31 days in December, and 28 of them contain 7 cycles, which are 7 days off. The remaining 3 days are to see when your first break appears?

Situation 1: Rest on the 1st, then count 28 days to the 29th, you should take 7 + 1 = 8 days, and go to work on the 30th and 31st;

Situation 2: Rest on the 2nd, then count 28 days to the 30th, also rest for 8 days, and go to work on the 31st;

Situation 3: Rest on the 3rd, then count 28 days to the 31st, and also rest for 8 days;

Situation 4: Rest on the 4th, then count 24 days to the 28th, rest for 7 days, and go to work on the 29th, 30th, and 31st;

The 5th day is the same as the situation 1, so the minimum day off is 7 days in December, and the maximum day off is 8 days.

You have taken 2 days off, leaving a minimum of 5 days of overtime and a maximum of 6 days of overtime.

3 days off 1 day, 29 days should be 9 days off, and then 2 days of work, did not meet the conditions of 3 days and 1 day off, so it is considered to be 7 days of overtime after 9 days of shift, 2 days of rest, so it should be 29 days of work, 7 days of overtime.

On 3 days off 1?? Isn't there a maximum number of days off per month? According to you, it should be about 7 days off. However, the specific situation needs to be analyzed on a case-by-case basis!

First find the area of the blank part in the lower left corner and the upper right corner, the area of the two parts is equal, both are:

Square area - circle area) 4 = (4 - 2) 4 = 4 - square centimeters.

Then find the area of a large blank part in the middle:

Draw a diagonal line to divide the blank part in the middle in two, so that each half of the block can be seen as the area of the arc minus the area of the triangle.

So the area of the middle blank part is.

4 4 (the arc is 1 4 circles) -1 2 4 (area of a right triangle)) 2 = 8 -16 square centimeters.

Then subtract the area of the empty space from the square area to find it:

4 -4-)2 - 8 -16) = 24 - 6 square centimeters.

The equation y= (1 2- 2x-x 2) for the blue of the great circle, the equation y= (1 4-x 2) for the regular semi-circumference of the small circle, the abscissa of point a is 2 8, the point b(1- 2 2,0), the point c(1 2,0), from a to c, the area under the small circle is (1 4-x 2)dx= 16- 7 64-(arcsin 2 4) 8, from a to b, the area under the great circle is (1 2- 2x-x 2)dx= 4-5 7 64-(arcsin5 2 8) 2, The subtraction of the two values is the area of the half crescent 7 16-3 16-(arcsin 2 4) 8+(arcsin5 2 8) 2, so the area of the crescent is 7 4-3 4-(arcsin 2 4) 2+2arcsin5 2 8

The area of the blank part in the lower left corner and the upper right corner is (4*4-4) 2=8-2, and the area of the blank part in the middle is the sum of the area of the two quarters, minus the square area 16, 4*2-16=8 -16

The sum of the areas of the blank parts is 6 -8

The sum of the shaded areas is 16-(6 -8) = 24-6

Solution: In the first step, the area of each of the four small corners = (the area of the square - the area of the middle circle) 4 = (16-4) 4 = 4-

Step 2, Area of the Sector = (4) 2 4=4 Step 3, Area of a Shadow = Area of Square-Fan-Small Corner = 12-3 Step 4, Area of Shadow = 2x(12-3) = 24-6

Blank part in the middle = 2 * (1/4 circle - isosceles right triangle) = 2 * (. 12

Blank space in both corners = 1 2 * (square - circle) = 1 2 * (. 72

then shaded area = . 16

Leaf shape blank in the middle: 1 4 4 4 - (4 4 - 1 4 4 4) = 8 -16

The remaining two blank spaces: 1 2 (4 4- 2 2) = 8-2 Therefore, the area of the shaded part: 4 4 - (8 -16) - (8-2 ) = 24-6 units square centimeter.

The upper left corner and the lower right corner are connected, 1, subtract the area of the last half of the square connection diagonal with the fan area, and then multiply by 2 the area of the middle blank space (denoted as a); 2. Subtract the area of the inscribed circle from the area of the square and divide it by 4 The area of the blank space in the lower left corner (denoted as b), the area of the square is 2b a shaded area.

First, find the area of two small corners: (square - circle area) * 1 2 = (4 * 4 - 2 * 2 * ) 1 2 = (16-4 ) * 1 2 = 8-2

Two large corners: 2 * (square area - 1 4 of a circle area) = (4 * 4 - 4 * 4 * * 1 4)) * 2 = 32-8

Shadow area: two large corners - two small corners = (32-8) - 8-2 ) = 24-6

Overlapping problem, the white block in the middle is the overlap of two semicircles, equal to one circle (radius 4) minus the square. The outside blank piece is a square minus a circle (diameter 4) divided by 2. The shadow is the square minus these two parts.

The area of the white part in the middle = 2 sector area - square area = 8pi - 16, and the blank area in the lower left corner = (square area - circle area) 4 = 4 - pi.

So shadow area = 16 - (8 pi - 16) - 2 (4 - pi) = 24 - 6 pi

First, calculate the area of the four corners and subtract the area of the circle from the square, i.e. (16-4) 4=4-

24-6 Add the area of the two quarter-semicircles and subtract the area of the square to get the white area in the middle. The square area minus the middle circle area and divide by four to get the white area in the upper right corner. Finally, subtract the white area in the middle with the square area, and then subtract the white area in the upper right corner and the lower left corner.

(1) Divide 35 by 7/11 to get 55

2) 720 divided by (1-4/19) to get 912

n×(n+1)/2

You can write and reverse the following numbers and find that the top and bottom are added up to n+1, there are a total of n terms, I hope you can understand.

n function you write y=x2, I default to n(x)=x, then there is:

dₓ=dₓ+x²

cₓ+cₓ)/2+x²

cₓ+cₓ+x²)/2+x²

cₓ+3x²/2

bₓ+bₓ)/2+3x²/2

bₓ+bₓ+x²)/2+3x²/2

bₓ+2x²

aₓ+aₓ)/2+2x²

aₓ+aₓ+x²)/2+2x²

aₓ+5x²/2

The question is given: a < a defines x as a natural number 1 8

So 8 is also the maximum.

So: d = d = a +5 8 2 = a +160

Next to the beam, the numerator and denominator are reversed.

If you are like this, c can be valued in the range r

If the numerator and denominator are reversed, i.e., the denominator x +2x-c cannot be taken as the minimum value of 0x+1) 2-1-c, and the minimum value of 1-c>0c<-1 is changed to -1-c, 1-c0c-1

x +2x-c is the denominator, right?

The denominator is always not 0, then.

It should be Shiwen 4+4c<0

c<-1

In this way, you can understand that the equation x +2x-c=0 does not have a positive return to the fast real number.

So b -4ac<0

4+4c<0

c<-1

Because y=(x.).+2x-c) 1, so y=x.+2x-c.

Since the disadvantage front takes any value for x, the formula has a world meaning, so 4-2 (-c-y) is greater than or equal to divination or 0

2 is greater than or equal to C+Y, so the range of values for C is less than or equal to 2-Y

Substituting y=kx+1 into y 2=2x gives us a quadratic equation about x. Then use Vedica's theorem to find x1+x2, x1x2, and substitute the equation. At this time, Paishan you can find the repentance k has two values.

But there is one to be discarded, because you have to make two solutions to the quadratic equation for x, i.e., its pre-dusty 0

Vaguely calculated, probably k=-3.

In the required formula, there is no y first, and Hu Hong eliminates y, and obtains )k 2) x 2+(2k-2)x+1=0

However, after the trace, use Veda's theorem).

x(1)+x(2)=(2-2k)/k^2

x(1)*x(2)=1/k^2

Bring it into the original style. 3-2k=k^2

k^2+2k-3=0

k=-3;k = 1 (remove the k that does not meet the range in (1) after the last book).