mk chưa học giai thừa ,xin lỗi nha

Cho hình bình hành ABCD. Gọi I,K theo thứ tự là trung điểm của CD, AB. Đường chéo BD cắt AI, CK theo thứ tự ở M và N. Chứng minh rằng:

a) AI // CK

b) DM = MN = NB

\(x^3+3x^2-3x-1=\left(x^3-1\right)+\left(3x^2-3x\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)

\(=\left(x-1\right)\left[\left(x^2+x+1\right)+3x\right]=\left(x-1\right)\left(x^2+4x+1\right)\)

1) x⁴y² + x²y⁴ + x⁴y³ + x²y⁵  = (x⁴y² + x²y⁴) + (x⁴y³ + x²y⁵) = x²y².(x² + y²) + x²y³.(x² + y²) = x²y².(x²+ y²) (1 + y) = [xy.(x² + y²)].[xy(1+y)]

=> x⁴y² + x²y⁴ + x⁴y³ + x²y⁵ chia cho xy.(x² + y²)  bằng xy.(1+ y)

2) A = (n² - 8)² + 36 = n⁴ - 16n² + 100  = (n⁴ + 20n² + 100) - 36n² = (n² + 10)² - (6n)² = (n² - 6n+ 10).(n² + 6n+ 10)

Vậy để A là số nguyên tố thì n² - 6n + 10 = 1 hoặc n² + 6n + 10 = 1

Mà n là số tự nhiên nên n²+ 6n + 10 > 1 

=>  n² - 6n + 10 = 1  => n² - 6n + 9 = 0 => (n -3)² = 0 => n = 3 

Vậy....

3) a) = xy(x - y) - xz(x + z) + yz.[(x+ z) + (x - y)] = xy(x - y) - xz(x + z) + yz.(x + z) + yz(x - y)

= [xy(x - y) + yz.(x - y)] + [(yz.(x+ z) - xz(x+z)] = y(x - y)(x+ z) + z(x + z).(y - x) = (x+ z)(x- y).(y - z)

b) = (x² + x)² - (2x)² - 4(x+3) = (x² + x + 2x).(x² + x- 2x) - 4(x+3) = (x² + 3x).(x² - x) - 4(x+3)

= (x+3).[x.(x² - x) - 4] = (x+3).(x³ - x² - 4) = (x+3).(x³ - 8 + 4 - x²) = (x+3).[(x - 2)(x² + 2x + 4) - (x - 2).(x+2)]

= (x + 3).(x - 2).(x² + 2x + 4 - x- 2) = (x + 3).(x - 2).(x² + x + 2) 

4) a) n⁴ + 1/4 = (n⁴ + n² + 1/4) - n² = (n² + 1/2)² - n² = (n² - n + 1/2).(n² + n + 1/2) = [n(n - 1) + 1/2].[n.(n+1) + 1/2]

Áp dụng công thức ta có:

A = \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right).\left(4^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}=\frac{\frac{1}{2}.\left(1.2+\frac{1}{2}\right).\left(2.3+\frac{1}{2}\right).\left(3.4+\frac{1}{2}\right)...\left(18.19+\frac{1}{2}\right).\left(19.20+\frac{1}{2}\right)}{\left(1.2+\frac{1}{2}\right).\left(2.3+\frac{1}{2}\right).\left(3.4+\frac{1}{2}\right).\left(4.5+\frac{1}{2}\right)...\left(19.20+\frac{1}{2}\right).\left(20.21+\frac{1}{2}\right)}\)

A = \(\frac{\frac{1}{2}}{20.21+\frac{1}{2}}=\frac{1}{841}\)