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\(\frac{2n+1}{\left[n\left(n+1\right)\right]^2}=\frac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(\Rightarrow C=\frac{1}{1}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(\Rightarrow C=1-\frac{1}{\left(n+1\right)^2}\)
$\textbf{a)}$
$A=\left(1-\dfrac12\right)\left(1-\dfrac13\right)\left(1-\dfrac14\right)\cdots\left(1-\dfrac1n\right)$
$=\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}$
$=\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}$
$=\dfrac1n.$
$\textbf{b)}$
$B=\left(1-\dfrac1{2^2}\right)\left(1-\dfrac1{3^2}\right)\cdots\left(1-\dfrac1{n^2}\right)$
$=\dfrac{(2-1)(2+1)}{2^2}\cdot\dfrac{(3-1)(3+1)}{3^2}\cdots\dfrac{(n-1)(n+1)}{n^2}$
$=\left(\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}\right)\left(\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{n+1}{n}\right)$
$=\dfrac1n\cdot\dfrac{n+1}{2}$
$=\dfrac{n+1}{2n}.$
\(N=\frac{-1^2}{1.2}.\frac{-2^2}{2.3}.\frac{-3^2}{3.4}....\frac{-100^2}{100.101}.\frac{-101^2}{101.102}\)
\(=\frac{1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}....\frac{100.100}{100.101}.\frac{101.101}{101.102}\)
\(=\frac{1.2.2.3.3....100.100.101.101}{1.2.2.3.3.4....100.101.101.102}\)
\(=\frac{1}{102}\)
( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0
<=> +) 1/7 . x - 2/7 = 0 +) (- 1 / 5) . x +3/5 = 0 +) 1/ 3 . x + 4/ 3 = 0
x = 2 x = 3 x = 4
Vậy x = 2 : x = 3 ; x=4
$\left(\dfrac17x-\dfrac27\right)\left(-\dfrac15x+\dfrac35\right)\left(\dfrac13x+\dfrac43\right)=0$
$\Leftrightarrow\dfrac{x-2}{7}\cdot\dfrac{3-x}{5}\cdot\dfrac{x+4}{3}=0$
$\Leftrightarrow x-2=0\ \text{hoặc}\ 3-x=0\ \text{hoặc}\ x+4=0$
$\Leftrightarrow x=2\ \text{hoặc}\ x=3\ \text{hoặc}\ x=-4.$
Vậy $\boxed{x\in\{-4,\ 2,\ 3\}.}$