$\textbf{a)}$

$A=\left(1-\dfrac12\right)\left(1-\dfrac13\right)\left(1-\dfrac14\right)\cdots\left(1-\dfrac1n\right)$

$=\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}$

$=\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}$

$=\dfrac1n.$

$\textbf{b)}$

$B=\left(1-\dfrac1{2^2}\right)\left(1-\dfrac1{3^2}\right)\cdots\left(1-\dfrac1{n^2}\right)$

$=\dfrac{(2-1)(2+1)}{2^2}\cdot\dfrac{(3-1)(3+1)}{3^2}\cdots\dfrac{(n-1)(n+1)}{n^2}$

$=\left(\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}\right)\left(\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{n+1}{n}\right)$

$=\dfrac1n\cdot\dfrac{n+1}{2}$

$=\dfrac{n+1}{2n}.$

#)Giải :

a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)

b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

$\textbf{A)}$

$A=2009^{(1000-1^3)}\cdot(1000-2^3)\cdots(1000-15^3).$

$\text{Vì }1000-10^3=1000-1000=0.$

$\Rightarrow A=0.$

Làm lại cho you đây -_- vừa nãy bấm mt nhầm,đời t nhọ vãi

1)\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+....+16\right)\)

\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+\dfrac{1+2+3+4}{4}+...+\dfrac{1+2+3+...+16}{16}\)

Xét thừa số tổng quát: \(\dfrac{1+2+3+...+t}{t}=\dfrac{\left[\left(t-1\right):1+1\right]:2.\left(t+1\right)}{t}=\dfrac{\dfrac{t}{2}\left(t+1\right)}{t}=\dfrac{\dfrac{t^2}{2}+\dfrac{t}{2}}{t}=\dfrac{t\left(\dfrac{t}{2}+\dfrac{1}{2}\right)}{t}=\dfrac{t}{2}+\dfrac{1}{2}\)

Như vậy: \(P=1+\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\left(\dfrac{3}{2}+\dfrac{1}{2}\right)+\left(\dfrac{4}{2}+\dfrac{1}{2}\right)+...+\left(\dfrac{16}{2}+\dfrac{1}{2}\right)\)

\(P=1+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+....+\dfrac{17}{2}\)

\(P=\dfrac{2+3+4+5+...+17}{2}\)

\(P=\dfrac{152}{2}=76\)

2) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)

\(\Rightarrow2016\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{2016}{a+b}+\dfrac{2016}{b+c}+\dfrac{2016}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{c+a}{c+a}+\dfrac{b}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=\dfrac{2016}{3}\)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{2016}{3}-1-1-1=\dfrac{2007}{3}\)

1,

\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2018}-1\right)\\ A=\left(-\dfrac{1}{2}\right)\cdot\left(-\dfrac{2}{3}\right)\cdot...\cdot\left(-\dfrac{2017}{2018}\right)\\ =-\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2017}{2018}\right)\\ =-\dfrac{1}{2018}\)