Math Emperor Soon, Math Boss, Come Soon?

Substituting y=m2 into the curve c1: y=x2 4, then a(-2m, m2), b(2m, m2), pa=pb=2m

Substituting y=m2 into the curve c2:y=x2 9, then c(-3m, m2), d(3m, m2), pc=pd=2m

Then ab=pa+pb=4m, cd=pc+pd=6m, no matter what value m takes, there is ab cd=2 3

The bottom of the triangle AOB is AB, the height is OP, the bottom of the triangle CQD is CD, the height is PQ, because O and Q are symmetrical with respect to the straight line AB, so OP=PQ, the area of triangle AOB is S1=1 2·AB·OP, the area of the triangle CQD is S2=1 2·CD·PQ, so the ratio of the area is AB CD=2 3

If the triangle AOB is an isosceles right triangle, then Pa=Op, Pa=2M, Op=M 2, then M=2, AB=4M=8, OP=M 2=4, the area of the triangle AOB is S1=1 2·ab·op=16, and the difference between the triangle COD and the triangle AOB is 1 2·S1=8

If the triangle CQD is an isosceles right triangle, then PC=PQ, PC=3M, PQ=OP=M 2, then M=3, CD=6M=18, PQ=M 2=9, the area of the triangle CQD is S2=1 2·cd·pq=81, and the difference between the triangle COD and the triangle AOB is 1 3·S2=27

Substituting x=-2m into the curve c2:y=x2 9, then the ordinate of e is 4 9·m 2, and ae=5 9·m 2

Substituting x=3m into the curve c1:y=x2 4, then the ordinate of f is 9 4·m 2, and df = 5 4·m 2

Because ae is parallel to the y-axis, the bottom of the triangle mae is ae=5 9·m2, the height is ap=2m, and the area is s3=1 2·ae·ap

In the same way, df is parallel to the y-axis, so the bottom of the triangle mdf is df=5 4·m2, the height is ad=3m, and the area s4=1 2·df·dp

So the area ratio of the triangle MAE to the triangle MDF is 8 27

Let me tell you a rough idea first: substituting y=m2 into the middle of C1 and C2 respectively to get xb=2m, that is, AB is 4m long, and CD is 6m long, so ** and blanks are filled.

After that, because op=pq, the area of the triangle AOB is 1 2·ab·op and the area of CQD is 1 2·cd·pq, and since op=pq, the area ratio is the ratio of the length of ab and cd.

Do you know the nature of isosceles right triangles? Assuming AOB is an isosceles right triangle, then op=ap=bpThen there is: 1 2·cd·pq-1 2·ab·op, because op=pq, can be integrated into 1 2·(cd-ab)·op

The third question is a little bit reluctant to do .........

Guess what, just guess a few.

ab/cd=2/3

2) 8 when AOB is a right triangle and 27 when CQD is a right triangle

This is a high school exam question, how do I remember that the parabola was a high school thing at that time, ah......Unable to ......

This is the question of the high school exam in which year.

a1=1,a2=2

a(n+2)-an=2 ;n is an odd number.

a(n+2)+1 = 3(an +1) n is an even number s20=?solution:

n is the odd number of things:

a(n+2)-an=2

an - a1 = n-1

an = n

n is the number of lead talks about the doll cover:

a(n+2)+1 = 3(an +1 )

is an equal envy ratio sequence, q=3

an +1 = 3^(n/2 -1) .a2+1)an = 1+ 3^(n/2)

iea(2n-1) =2n-1

a1+a3+..a19 = 5(19+1) =100a(2n) =1+ 3^n

a2+a4+..a20

s20a1+a2+..a20

ans : a

1.From the question, it can be seen that the length of the two right-angled sides is 8 and 6 respectively, so the area is 8*6 2=242The circumference of an equilateral triangle is 18, so the length of the base edge is 18 3=6, and the area is 6*4 2=12

3.Both are equal in size.

That is: base (parallelogram) * height (parallelogram) base (triangle) * height (triangle) 2

The bases of the two are equal, so: high (parallelogram) high (triangle) 2 high (triangle) 2*high (parallelogram).

It represents the multiplication sign and the division sign.

Hope it helps....

Because they are all positive numbers, you can directly remove the absolute value, and the middle number is combined in pairs to get 0

Finally = 1 10-1 50 = 4 50 = 2 25

470 60 + 90 + 150 + 60 + 60 + 60 (this is the most difficult to make up, because the difference between these approximations is a multiple of 30, and the approximation is more than 470, but it is actually smaller than 470).

Among these five approximate numbers, only the first one is long, and the rest of the approximations are smaller than the original numbers, and the above may not be the closest, sometimes you have to use 30 as many times as possible, but don't exceed it.

Find it with a derivative.

f（x）=x³-3ax-1

f'（x）=3x²-3a=3（x²-a）

1. When a 0, f(x) is constant.

2, a is greater than 0, x a, that is, when x root a or x is less than - root a, f'(x) 0, f(x) increments.

x a, i.e., when -root a x root a, f'(x) 0, f(x) decreasing. Hope.

Derivative y'=3x 2-3a=3(x 2-a) When the above equation is 0, the function increases, and when it is < 0, the function decreases.

a Discussion on a basis.

When a<0, since x 2 is equal to 0, y'Constant》0 At this point, the function increases monotonically.

When a>0, x 2-a has two roots, i.e. (-root number a) when the root number a y'>0 function increments.