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

a) \(A=\left(\sqrt{x}+3\right)^2-4\sqrt{x}-6\)

\(A=x+6\sqrt{x}+9-4\sqrt{x}-6\)

\(A=x+2\sqrt{x}-3\)

b) \(A=x+2\sqrt{x}-3\)

\(A=x+3\sqrt{x}-\sqrt{x}-3\)

\(A=\sqrt{x}\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)\)

\(A=\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)\)

a: A=x+6căn x+9-4căn x-6

=x+2căn x+3

b: A ko phân tích được nha bạn

3{x-2}²/x{x-2}{x²+2x+4}=3{x-2}/x²+2x+4=-3/x+2 

nho tick minh nha

Bài 6:

c: \(9x^2+6x+1=\left(3x+1\right)^2\)

d: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)

e: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

a, \(=\left(x+y\right)^2-2^2=\left(x+y-2\right)\left(x+y+2\right)\)

b, =  \(y^2-4x^2+4x^2=y^2\)

Thay y = 10 vào BT trên, ta có:

\(y^2=10^2=100\)

Vậy giá trị của BT là 100

b) 2xy(1+z)

d) 33x^2

f) (x-2y)(3x+y)

b, 2xy+2xyz
=2xy(1+z)

d,27x^2+6x^2
=3x²(9+2)
=3x².11

f, (3x-6y)x+y(x-2y)
=3(x-2y)+y(x-2y)
=(3+y)(x-2y)

b: \(2xy+2xyz=2xy\left(z+1\right)\)

d: \(27x^2+6x^2=33x^2\)

f: \(\left(3x-6y\right)x+y\left(x-2y\right)=\left(x-2y\right)\left(3x+y\right)\)