CMR với mọi số nguyên n thì \(n^3+5n⋮6\)
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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, \(x^2-4y^2+1+2x=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)
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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 \(A=90\)
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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
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\(49-4x^2+2x+7\)
\(=-4x^2+2x+56\)
\(=-2\left(2x^2-x-28\right)\)
\(=-2\left(2x^2-8x+7x-28\right)\)
\(=-2\left[2x\left(x-4\right)+7\left(x-4\right)\right]\)
\(=-2\left(x-4\right)\left(2x+7\right)\)
\(49-4x^2+2x+7=-4x^2+2x+56\)
\(=-2\left(2x^2-x+28\right)=-2\left(2x^2-8x+7x+28\right)\)
\(=-2\left[2x\left(x-4\right)+7\left(x-4\right)\right]=-2\left(2x+7\right)\left(x-4\right)\)
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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 \(⋮6\forall n\inℤ\)