c: Sửa đề: \(n^2+9\) ⋮n-1

Ta có: \(n^2+9\vdots n-1\)

=>\(n^2-n+n-1+10\vdots n-1\)

=>10⋮n-1

=>n-1∈{1;-1;2;-2;5;-5;10;-10}

=>n∈{2;0;3;-1;6;-4;11;-9}

d: \(2n^2+n+7\vdots n+2\)

=>\(2n^2+4n-3n-6+13\vdots n+2\)

=>13⋮n+2

=>n+2∈{1;-1;13;-13}

=>n∈{-1;-3;11;-15}

\(2^3.2^4=2^{3+\text{4}}=2^7\)

khó.

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Ok

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c: \(2448:\left[199-23-6\right]\)

\(=2448:90\)

\(=\dfrac{136}{5}\)

a) \(5^6:5^4+2^3.2^2-1^{2017}\)

\(=5^2+2^5-1=25+32-1=56\)

a) 80-(4x5^2-3x2^3)

   = 80-(4x25-3x8)

   = 80-(100-24)

   = 80-76

   = 4

b) 125-2x[56-48:(15-7)]

  = 125-2x[56-48:8]

  = 125-2x[56-6]

   =125-2x50

   =125-100

   =25.

2448:[119-(23-6) ]

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

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