Ta có: \(\frac{11}{2}.\frac{12}{2}.\frac{13}{2}....\frac{20}{2}=\frac{11.12.13....20}{2^{10}}=\frac{11.13.15....19.\left(12.14.16.18.20\right)}{2^{10}}=\)

\(=\frac{11.13.15....19.\left(\left(3.2^2\right).\left(7.2\right).\left(2^4\right).\left(9.2\right).\left(5.2^2\right)\right)}{2^{10}}=\frac{11.13.15....19.\left(3.5.7.9\right).2^{10}}{2^{10}}=\)

\(=1.3.5.7.9.11.13.15....19\left(Đpcm\right)\)

Để A là số nguyên thì \(n-1⋮n+4\).

Ta có : n - 1 = (n + 4) - 5

Do n + 4 \(⋮\)n + 4

Để (n + 4) - 5\(⋮\)n + 4 thì 5 \(⋮\)n + 4 => n + 4\(\in\)Ư(5) = {1; -1; 5; -5}

Lập bảng : 

Vậy n = {-3; -5; 1; -9} thì A là số nguyên

Gọi d=ƯCLL(n-1;n+4)

Suy ra n-1chia hết cho d

           n+4_________d

a, = 1

b 1 = S 

\(T=\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\)  => \(\frac{T}{2}=\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\)

=> \(T-\frac{T}{2}=\left(\frac{2}{2^1}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2017}{2^{2016}}\right)-\left(\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{2017}{2^{2017}}\right)\)

<=> \(\frac{T}{2}=\frac{2}{2^1}+\left(\frac{3}{2^2}-\frac{2}{2^2}\right)+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{2017}{2^{2016}}-\frac{2016}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)

<=> \(\frac{T}{2}=1+\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)-\frac{2017}{2^{2017}}\)

Đặt: \(M=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}=>2M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\)

=> \(2M-M=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)

=> \(M=\frac{1}{2}-\frac{1}{2^{2016}}< \frac{1}{2}\)

=> \(\frac{T}{2}< 1+\frac{1}{2}-\frac{2017}{2^{2017}}< 1+\frac{1}{2}=\frac{3}{2}\)

=> T < 3

\(\frac{x+2}{2}=\frac{8}{x+2}\)

\(\Rightarrow\left(x+2\right)\cdot\left(x+2\right)=8\cdot2\)

\(\Rightarrow\left(x+2\right)^2=16\)

\(\Rightarrow\left(x+2\right)^2=\left(\pm4\right)^2\)

\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)

vì x+2

_____________________=8___________

2                                        x+2

=>(x+2)(x+2)=8.2

=>(x+2)^2=16

=>(x+2)^2=4^2

=>x+2=4 hoặc -4

nếu x+2=4=>x=2

nếu x+2=-4=>x=-6

vậy x thuộc{4;-6}

k cho mk nha

Đặt  A  =\(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)    

Ta có \(3A=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

           \(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

     => \(2A=3A-A=3-\frac{1}{3^{2005}}\)

   => \(A-\frac{3-\frac{1}{3^{2005}}}{2}\)

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