Who would do this annoying math problem, hurry up, I m in a hurry!

Solution: f 1 2008 = 1 2008 1 1 2008 =1 2009, f 2008 = 2008 1 2008 = 2008 2009;

So f 1 2008 +f 2008 =1 2009 + 2008 2009 = 1;

In the same way, f 1 2007 +f 2007 =1 2008 + 2007 2008 = 1;

f﹙1／2006﹚+f﹙2006﹚=1／2007+2006／2007=1；

So f 1 2008 f 1 2007 f 1 2006f(1／3﹚＋f﹙1／2﹚＋f﹙1﹚＋f﹙1﹚＋f﹙2﹚＋f﹙3﹚＋.f﹙2006﹚＋f﹙2007﹚＋f﹙2008）=[f﹙1／2008﹚＋f﹙2008）]+f﹙1／2007﹚＋f﹙2007﹚]＋f﹙1／2006﹚＋f﹙2006﹚]＋f(1／3﹚＋f﹙3﹚＋f﹙1／2﹚＋f﹙2﹚+f﹙1﹚＋f﹙1﹚=1+1+1...

In fact, there are many similar problems in mathematics, mainly to examine the idea of before and after the combination, and you can consider this way of thinking when you see this very complex formula, I hope it will inspire you.

f(1/2008)+f(2008)=1

f(1/2007)+f(2007)=1

Then there are a total of 2008*2 items here.

The sum of the two items is equal to 1

Thus the answer is equal to 2008

Defining the domain is all real numbers, that is, when x takes any value, y is meaningful, then this problem can be started by the y function is meaningful, it is easy to know, for y, if and only if (mx2+4x+m+2) is evergreen in 0, y is meaningful, thus, then this problem becomes: when m is the range of values, the function z=(mx2+4x+m+2) is evergreen in 0.

In this case, if any value of x (mx2+4x+m+2) is greater than 0, that is, y is meaningful, then the domain is defined as all real numbers.

So how do you find m? Observing the function z=mx2+4x+m+2, it can be seen that m is the coefficient of x2, then it is necessary to divide whether m is 0, obviously, if m is 0, then the function z cannot be as large as 0, so m can not be 0, it is easy to know that the function z opening has two kinds: up and down, if down, that is, m is less than 0, then there must be a function curve under the x-axis, that is, there is x so that z is less than 0 (the general function diagram is drawn on the papyrus, you can know it at a glance), so m should be greater than 0, and then look at z=mx2+4x+m+2. It can be seen that at this time, if and only if the equation mx2+4x+m+2=0 has no real root (that is, when there is no intersection with the x-axis), Z Evergrande is 0 (a drawing can be known), there is an equation mx2+4x+m+2=0 to calculate that the discriminant formula should be less than 0, that is, m2+2m-4 should be greater than 0, solve this inequality, obtain the value range of m, and m is greater than 0 to take the intersection...

(mx2+4x+m+2) to the minus 1 to the 4th power.

To be meaningful, then (mx2+4x+m+2) must be greater than 0 to define the domain as a whole real number, which means that no matter what value x takes, mx 2+4x+m+2>0 is constant.

Let y=mx 2+4x+m+2

Then apparently this is a parabola.

If you want mx 2+4x+m+2 to always be greater than 0

Then the parabola should be above the x-axis.

So the parabolic opening must be upward, m>0

At the same time, the parabola cannot have an intersection point with the x-axis, otherwise, there will always be one or more or infinite x-making, mx 2+4x+m+2<=0, so the equation y=mx 2+4x+m+2 should not have a real root, then 4 2-4m(m+2)<0

m>-1+ 5 or m<-1-5

Due to m>0

So finally. m>-1+√5

To say that defining the domain as a whole real number is to ask you how much m is, mx2+4x+m+2 is evergreen to equal zero; Extract m.

m(x2+4x/m+1+2/m)=m[(x+2/m)^2-4/m^2+1+m/2];DiscussionWhen m>0, the minimum value of y=(x+2 m) 2-4 m 2+1+m 2 should also be greater than or equal to zero; When m<0, the maximum value of y=(x+2 m) 2-4 m 2+1+m 2 should also be less than or equal to zero; Finally, the interval can be merged.

Defining the domain as all real numbers, indicating that the thing below 1 4 squares is forever "0", that is, mx2+4x+m+2 is always greater than zero, that must be a parabola that does not intersect with the x-axis and has an upward opening. So m>0, <0

16-4m*(m+2)<0

Get m> root number 5-1

Laid daily.

The campus network is long.

300 3 20 2000 meters.

Good luck with your studies!

Hope it helps.

Thank you!

If the worker lays 60% of the total length in 4 days, then one day is 60% of the total length 4=15%.

Then, for a total of five days, it is 5*15%=75% of the total length, which is 300 meters.

So, the overall length is 300 75% = 400 meters.

2000m

One day is 300m, four days is 1200m, and 1200m is 60% of the total length

The coordinates of point P are (5,3) and the coordinates of point A are (1,0).

The parabolic analytic formula is: y=(-3 16)(x-5) 2+3

Analysis: There is such a theorem in the circle: the radius of the tangent point connected by the number must be perpendicular to the tangent, so it is easy to get that the abscissa of the point p is 5, and the ordinate of the solution is 3 by substituting y=(3 5)x, connecting AP, and passing the point P as the perpendicular line of Sakura AB intersects AB in E, the hypotenuse ap that constitutes a right triangle is the radius 5, the right-angled side PE is the ordinate of the P point 3, and the crack solution is AE 4, and because the length of OE is 5, the coordinate of point A can be obtained as (1,0).

Since the axis of symmetry of the parabola is x=5, it can be assumed that the parabolic equation is y=a(x-5) 2+b, and the coordinates of the points p and a are substituted into the solution a=-3 16 b=3 respectively

Do not pass through point D.

Analysis: First, the coordinates of point c (0,3) are obtained, then the point d coordinates of the symmetry about the origin point are (0,-3), and x=0 is substituted into the parabolic equation to solve y=-27 16≠-3, so point d is not on the parabola.

Exist. The parabola in the problem passes through a, b, and c and the vertex is on the straight line y=(3 5)x, in fact, the required line i is y=(3 5)x(x 0).

The actual distance between A and B is 180 km, and the distance between A and B is 6 cm on a map. The scale is: 1:3000000.

If it is on a map with a scale of 1:5000000, the actual distance between A and B is 180 kilometers, and A and B should be drawn in centimeters.

If A and B are separated by xkm, the speed of the express train is ykm h, and the speed of the slow train is five-sevenths of ykm h.

According to the title:

x=4(y+5 7y) [The fast train and the slow train add up to four hours to meet, that is, they add up to the whole distance.] 】

x/2-48=20/7y

Local. 】x/2+48=4y

Express. ] can be solved by linkage.