a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)

\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)

mk cảm ơn ah

\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\\ \Rightarrow A=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\\ \Rightarrow A=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{16}{2}+\dfrac{17}{2}\\ \Rightarrow A=\dfrac{1}{2}\left(2+3+4+...+17\right)=76\)

M=M mà M phải tick mk mới cho biết

sao ban lai xin loi 

Lời giải:
$1+\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+...+\frac{1}{16}(1+2+3+...+16)$

$=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{16}.\frac{16.17}{2}$

$=1+\frac{3}{2}+\frac{4}{2}+....+\frac{17}{2}$

$=\frac{2+3+4+...+17}{2}=\frac{1+2+3+...+17}{2}-\frac{1}{2}=\frac{17.18}{2.2}-\frac{1}{2}=76$

xét: Sn = 1 + 2 + 3 + 4 + ... + n (1) 
=> Sn = n + (n-1) + .. + 2 + 1 (2) 
thấy 1+n = 2 + (n-1) = 3+(n-2) = n-1 + 2 = n+1 
lấy (1) + (2) và với chú ý trên ta có: 
2.Sn = (n+1) + (n+1) +..+ (n+1) = n(n+1) (vì n số hạng giống nhau) 
=> Sn = n(n+1)/2 => Sn /n = (n+1)/2 

=> P = 1 + S2/2 + S3/3 + S4/4 +...+ Sn /n 

P = 1 + 3/2 + 4/2 + 5/2 +.. + (n+1)/2 

P = 2(2 + 3 + 4 + ... + n + n+1) = 2(1+2 +..+ n+1) - 2 = 2.S(n+1) - 2 

P = 2.(n+1)(n+2)/2 - 2 = (n+1)(n+2) - 2 = n²+3n 

Bài toán chỉ tính đến S16/16 (tức n = 16) 
P = 16² + 3.16 = ...

xét: Sn = 1 + 2 + 3 + 4 + ... + n (1)
=> Sn = n + (n-1) + .. + 2 + 1 (2)
thấy 1+n = 2 + (n-1) = 3+(n-2) = n-1 + 2 = n+1
lấy (1) + (2) và với chú ý trên ta có:
2.Sn = (n+1) + (n+1) +..+ (n+1) = n(n+1) (vì n số hạng giống nhau)
=> Sn = n(n+1)/2 => Sn /n = (n+1)/2

=> P = 1 + S2/2 + S3/3 + S4/4 +...+ Sn /n

P = 1 + 3/2 + 4/2 + 5/2 +.. + (n+1)/2

P = 2(2 + 3 + 4 + ... + n + n+1) = 2(1+2 +..+ n+1) - 2 = 2.S(n+1) - 2

P = 2.(n+1)(n+2)/2 - 2 = (n+1)(n+2) - 2 = n²+3n

 bài toán chỉ tính đến S16/16 (tức n = 16)
P = 16² + 3.16 = ...

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\).

\(A=1+\frac{1}{2}.3+\frac{1}{3}.6+....+\frac{1}{16}.136\)

\(A=1+1,5+2+...+8,5\)

\(A=\frac{\left(8,5+1\right)\left[\left(8,5-1\right):0,5+1\right]}{2}=76\)

A = 1 + 1/2 x 2x3/2 + 1/3 x 3x4/2 +.............+ 1/16 x 16x17/2

A = 1+ 3/2 +4/2 + ............+ 17/2

A = 1+ (3+4+5+.........+17)/2

A = 1+75 = 76