A math problem for the second year of junior high school! 100,000 urgent!!

The 5th equation is:

The k-th equation is.

1^3+2^3+3^3+4^3+5^3+..k^3=(1+2+3+··k)^2

Proof: The k-th equation is known to be.

1^3+2^3+3^3+4^3+5^3+..k^3=(1+2+3+··k)^2

So, the k+1 equation is.

1^3+2^3+3^3+4^3+5^3+..k^3+（k+1）^3=(1+2+3+··k)^2+（k+1）^3

2+（k+1）^3

1+2+3+··k+k+1)^2

So, the k+1 equation also holds.

The 5th equation is:

The k-th equation is 1 + 2 +k³=(1+2+3+..k)²

The first k+1 equation = [ k(k+1) 2] +k+1) =1 4[(k+1) +k +4k+4)]=1 4(k+1) (k+2) =(1+2+.k+k+1)²

1^3+2^3+3^3+4^3+..k^3=(k(k+1)/2)^2

Proof : Let k=k+1 be brought into the above equation to prove.

2. 1³+2³+3³+…k-1) +k =[k(k+1) 2] Prove: The k+1 equation is.

1³+2³+3³+…k-1)³+k³+(k+1)³(k(k+1)/2)²+k+1)³

k+1)²*k/2)²+k+1)³

k+1)²*k/2)²+k+1)]

k+1)²*k²/4+k+1]

k+1)²*k/2+1)²

k+1)²*k+2)/2]²

k+1)(k+2)/2]²

It's hard to write... Wish.

With the Pythagorean theorem.

Under the root number (x 2 + 120 2) = 200 2

x=160, so the answer should be (300-160): 20=7 seconds.

Because eb=ec. So the angle EBC = angle ECB, because the angle ABE is equal to the angle ACE, so the angle ABC= angle ACB, so the edge AB=AC, and because EB is equal to EC, the angle ABE is equal to the angle ACE, so the triangle ABE is congruent with the triangle ACE, so the angle BAE is equal to the angle CAE

This question seems to be based on basic training.

Figure is not clear, eb=ec

EBC = ECB (equilateral equilateral equiangular angle).

and d is the midpoint of BC.

abe= ace (known).

ABC = ACB (Equal plus Equal).

AD bisects BAC (isosceles triangle three lines in one).

6(1/x+1/y)=1 (1/x+1/y)=1/6

10(1/y+1/z)=1 (1/y+1/z)=1/10

5(1/x+1/z)=2/3 (1/x+1/z)=2/15

The first formula minus the second formula 1 x-1 z=1 15 and the third formula add 2*(1 x)=3 15=1 5

1/x=1/10 x=10

1/y=1/6-1/x=1/6-1/10=1/15 y=151/z=2/15-1/x=2/15-1/10=1/30 z=30

This topic is not difficult, it's just that something like 1 x scares you.

Just replace 1 x, 1 y, and 1 z with a, b, and c respectively to become a ternary system of equations.

That is, 6a+6b=1

10b+10c=1

5a+5c=2/3

Solution-available. a=1/10,b=1/15,c=1/30

So find the reciprocal.

x＝10,y=15,z=30

From the first equation we can know 1 x = 1 6-1 y, from the second formula we can know 1 z = 1 10-1 y, and then bring these two formulas into the third equation, 5 (1 6-1 y + 1 10-1 y) = 2 3, and calculate 1 y = ?, and then bring this value to the above two formulas, and find them out.

Simplify the three equations to get 1 x+1 y=1 6 1 y+1 z=1 10 1 x+1 z=2 15

Then add the three equations to 1 x+1 y+1 z=1 5, and subtract the first with the fourth equation.

I. 2. The third equation can be obtained to obtain how much 1 x 1 y 1 z is equal to, so that the values of x, y, and z can be obtained.

Let's as, 1 x=a 1 y=b 1 z=c6(a+b)=1 a=1 6-b10(b+c)=1 c=1 10-b5(a+c)=2 3 a+c=2 15 a=1 10 b=1 15 c=1 30x=10 y=15 z=30 Almost.

Transform the form to obtain: 1 x+1 y=1 6 ; The sum of the 3 formulas yields: 2 (1 x+1 y+1 z)=1 6+1 10+2 15;

1/y+1/z=1/10;So 1 x+1 y+1 z=1 5

1/x+1/z=2/15;Subtract the first few with this equation to get 1 z=1 30 z=30;

Subtracting the second equation from this equation gives 1 x=1 10 x=10;

The same goes for y=15

1) Pass point A to BC as AE BC, and pass MN to point FMN BC AMF ABP

mf/bp=af/ap

i.e. x-3 5=(6-ef) 6

ef=(48-6x)/5

s=x*(48-6x)/5

(6x -48x) 5]=-(6 5)x +(48 5)x2) when mn=-b 2a=4 has a maximum value of (4ac-b) 4a=96 5

Make AE perpendicular to the bottom and E on BC.

ae=6be=8-3=5

The triangle BPM is similar to the triangle BEA

bp/be=mp/ae

bp=5x/6

pc=8-5x/6

s=mp*pc

x(8-5x 6) and 0

Solution: connect od, oe, o and the extension line and bc edge of ab and ac in abc tangent, af=ad,be=bf,ce=cd,od ad,oe bc, acb=90°,quadrilateral odce is square, let od=r, then cd=ce=r,bc=3,be=bf=3-r,ab=5,ac=4,af=ab+bf=5+3-r,ad=ac+cd=4+r,5+3-r, r=2, then the radius of o is 2

So the answer is: 2