It s so difficult, 1 math problem 1, master help, 11

1.The 70th power of 6 is equal to the 35th power of 6 squares, which is the 35th power of 36, so it is larger than the 35th power of 35.

2.The 99th power of 9 is equal to the 9th power of 9 multiplied by the 9th power of 11.

99 to the 9th power is equal to 9 to the 9th power multiplied by 9 to the 90th power.

The numerator and denominator are about 9 to the 9th power, which is equal to the 9th power of 9 to the 90th power of 11.

3.The end number of 2000 to the power of 1999 is 0 because the end number of 2000 is 0 to the power of 01999 and the end of the power of 2000 is 1, because the even power of 9 is 1 and the odd power is 9

Question 1 35 to the 35th power, we will temporarily regard it as the 35th power of 6x6, that is, the 70th power of 6, and because it is 36>35, the 70th power of 36 is larger than the 35th power of 35.

In question 2, if you divide the 99th power of 99 by the 9th power of 9 by the 9th power of 9, you will find that it is exactly equal to the 9th power of 9 to the 90th power of 11.

Question 3 you can first observe that 9x9=81, the end is 1, 1x9=9 is 9 at the end, I don't know if you found that there is no number ending with 9 must be 1 or 9 at the end of the nth power, and we see that the end of 2 9s multiplying is 1, 3 9s multiplying the end is 9, and so on 1999 to the 2000th power at the end of 1, as for 2000 to the 1999th power is simpler, it is not difficult to see from the above that the end must be 0, 0x0 = 0

1+2+……n=n(n+1)/2

>1/(1+2+……n)=2/n(n+1)=2[1/n - 1/(n+1)]

> Original = 2*[1-1 2 + 1 2-1 3 +....1/99-1/100 + 1/100-1/101]

Split term method, very common.

It's probably high school plane analytic geometry, so I'll use a version suitable for the exam to solve the problem.

Set the point P coordinates to (A, A-2).

then kpa=(a-3) (a+1),kpb=(a-3) (a-1)tg apb=(kpa-kpb) (1+kpb*kpa)=(3-a) (a 2-3a+4).

Let a-3=b, then ctg apb=-(3+b+4 b)ctg apb>=1 (b=-2 take the equal sign) or ctg apb<=-7 (b=2 take the equal sign).

i.e. -1 7<=tg apb<=1

The equal signs on both sides are obtained as a=5 and a=1 respectively.

So, when a=1, i.e., p(1,-1), apb=4 is maximum.

Hope it helps.

p（1,-1） 45°

Put abp into the triangle and think, this is not in the line of the three points must have an external garden, and the chord ab has been determined, then the smaller the circle, the greater the circumferential angle corresponding to the chord ab, when the circle is tangent to the straight line, the radius is the smallest, at this time the distance from p to a and b in the center of the circle is equal to the distance from p to the straight line, p just coincides with the origin of the coordinates o.

In fact, the problem is very simple, you say that the point you are looking for is set c(x, x-2), and then combined with a(-1 1), b(1 1) can calculate the slope containing x, and then use the angle formula to express the angle containing x.

Then it's a matter of finding the best value!

I don't know if I'm satisfied, good luck!

Calculate 8 9 + 16 225 + 24 1225 + 32 3969 + 40 9801 + 48 20449 in a similar way

There should be a process, be detailed.