Two math verifications 20 added .

Evidence: 1) Write dbc= , dcb= , have the sum of the inner angles of the triangle to be 180°, + d=180°

2α+2β+∠a=180°

Synthesis, i.e., d=90°+1 2 a 2) mark abc= , acb= , cbf= , bcf= , with the sum of the inner angles of the triangle is 180°, +f=180°

+a=180°

And the flat angle is 180 °, there is.

Combining these four formulas, we get f=90°-1 2 a

Solution: (1) 2 3 x 4 0

2/3 x＝4

x＝－63x＋a＝0

3x＝－ax＝－a/3

a/3－（－6）＝2

a/3＋6＝2

a/3＝－4

a＝122）∵|a-3|0, (b 1) 0 again: |a-3|＋（b-1）²＝0

a－3＝0 a＝3

b－1＝0 b＝1

From the inscription: (2b a m) 2 (b 2a m) 1(2 3 m) 2 (1 6 m) 1

m＋1）/2＝7/6

m＝－10/3

3）①3kx－6＝（k＋3）x

3k－k－3）x＝6

2k－3）x＝6

x＝6/（2k－3）

The solution of this equation is a positive integer, and (2k 3) must be a positive factor of 6, which can only be .

2k－3＝1 k＝2

2k－3＝2 k＝

2k－3＝3 k＝3

2k－3＝6 k＝

Again: k is also a positive integer, k 2 or k 3

When k 2, x 6

When k 3, x 2

|1 1 1 ..1 |

a1 a2 a3 ..an |

a1^2 a2^2 a3^a ..an^2|

.= da1^(n-1) a2^(n-1) a3^(n-1) .an^(n-1)|

Such a determinant is the Vandermond determinant, and the result is: ii(ai-aj).

1<=j ('<=' means less than or equal to, 'ii' means multiplication).

Also, the sufficient necessary conditions for Vandermund's determinant to be zero are a1, a2, a3....n, n numbers, at least two equal

Let the general term of the Fibonacci sequence be an.

In fact an = (p n - q n) 5, where p = ( 5 - 1) 2, q = ( 5 + 1) 2. But it is not necessary to solve it here)

Then write SN = A1 + A2 + AN

Since an = sn - s(n-1) = a(n-1) +a(n-2) = s(n-1) -s(n-2) +s(n-2) -s(n-3).

s(n-1) -s(n-3)

The initial values are s1 = 1, s2 = 2, and s3 = 4.

So sn - 2s(n-1) +s(n-3) = 0

Thus its characteristic equation is.

x^3 - 2x^2 + 1 = 0

i.e. (x - 1)(x 2 - x - 1) = 0

It is not difficult to solve this cubic equation.

x1 = 1

x2 = p

x3 = q

p, q values are the same as p, q in an).

So the general solution is.

sn = c1 * x1^n + c2 * x2^n + c3 * x3^n

The values of C1, C2 and C3 are determined by substituting the three initial values of S1, S2 and S3 into the above equation. I won't forget it.

How do you know that the upstairs is square, and the pool should be rectangular.

Since it is a phalanx, then it is advisable to set this phalanx to be a phalanx of m*n, and the phalanx of (m-8) (n-8) is dug out in the middle.

The innermost layer is:

m-6)(n-6)-(m-8)(n-8)=64, then there is m+n=46

The outermost layer is:

mn-(m-2)(n-2)=2(m+n)-4=88 in total. mn-(m-8)(n-8)=8(m+n)-64=304

Each time you add a middle layer, the number of willows on each side increases by 2.

There are willows on each side of the innermost layer: 64 4 + 1 = 17.

So there are willows on each side of the outermost layer: 17 + 2 + 2 + 2 = 23.

There are willows in the outermost layer: (23-1) 4=88.

The penultimate layer has 17 + 2 = 19 trees on each side.

The penultimate layer has willows: (19-1) 4=72.

The penultimate layer has trees on each side: 19 + 2 = 21 trees.

The penultimate layer has willows: (21-1) 4=80.

There are a total of willow trees in the phalanx: 64 + 72 + 80 + 88 = 304 trees.

The triangular ADM area is 1 4 of the parallelogram area, the trapezoidal BMCD area is 3 4, the triangle DEM area is 2 times the BEM area, the triangle CDE area is 4 times the triangle BEM area, and the shaded area is 1 3

m is the midpoint of ab.

BM is 1 to 2 than CD, while triangle MBE is similar to triangle CDE De ratio to BE=2 to 1

And the triangle cdb=

Triangle ceb = 1 6

Area of the shaded part = 1 3