Ta có n³ + 5n

= n³ - n + 6n

= n(n² - 1) + 6n

= (n - 1)n(n + 1) + 6n

Vì (n - 1)n(n + 1) là tích 3 số nguyên liên tiếp

=> (n - 1)n(n + 1) \(⋮\)6

mà 6n \(⋮\)6

=> (n - 1)n(n + 1) + 6n \(⋮\)6

=> n³ + 5n \(⋮6\forall n\inℤ\)

a, \(2x^2-4xy+2y^2-32=2\left(x^2-2xy+y^2-16\right)\)

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

b, \(x^2-16+y^2+2xy=\left(x+y\right)^2-16=\left(x+y-4\right)\left(x+y+4\right)\)

c, \(x^2-4y^2+1+2x=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)

Ta có x = 99

=> x + 1 = 100

Khi đó A = x⁵ - 100x⁴ + 100x³ - 100x² + 100x - 9

= x⁵ - (x + 1)x⁴ + (x + 1)x³ - (x + 1)x² + (x + 1)x - 9

= x⁵ - x⁵ - x⁴ + x⁴ + x³ - x³ - x² + x² + x - 9

= x - 9 

Thay x = 99 vào A 

=> A = x - 9 = 99 - 9 = 90

Vậy A = 90

Ta có : \(x=99\Rightarrow100=x+1\)

\(A=x^5-100x^4+100x^3-100x^2+100x-9\)

\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-9\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-9\)

\(=x-9\)hay \(99-9=90\)

Vậy \(A=90\)

Giúp tui câu b đi 😢

a, Với m = 3 ta được : 

<=> \(f\left(x\right)=2x^3+5x^2+5x+3\)

Ta có : \(f\left(x\right)⋮h\left(x\right)\)hay \(2x^3+5x^2+5x+3⋮x+1\)

b, 

Để m - 2 = 0 <=> m = 2

\(49-4x^2+2x+7\)

\(=-4x^2+2x+56\)

\(=-2\left(2x^2-x-28\right)\)

\(=-2\left(2x^2-8x+7x-28\right)\)

\(=-2\left[2x\left(x-4\right)+7\left(x-4\right)\right]\)

\(=-2\left(x-4\right)\left(2x+7\right)\)

\(49-4x^2+2x+7=-4x^2+2x+56\)

\(=-2\left(2x^2-x+28\right)=-2\left(2x^2-8x+7x+28\right)\)

\(=-2\left[2x\left(x-4\right)+7\left(x-4\right)\right]=-2\left(2x+7\right)\left(x-4\right)\)

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