đề cụt ngủn vậy luôn ?

$\textbf{a)}$

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

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

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

$=\dfrac1n.$

$\textbf{b)}$

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

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

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

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

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

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)\)

=>  \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{100^2}\right)\)

     \(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{100^2-1}{100^2}\)

     \(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)

     \(=\frac{1.2....99}{2.3....100}.\frac{3.4....101}{2.3....100}\)

     \(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

=>  \(A=-\frac{101}{200}< -\frac{1}{2}\)

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

$\textbf{A)}$

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

$\text{Vì }1000-10^3=1000-1000=0.$

$\Rightarrow A=0.$

a)

Điều kiện: $x\ne-2,-5,-10,-17.$

$\dfrac3{(x+2)(x+5)}+\dfrac5{(x+5)(x+10)}+\dfrac7{(x+10)(x+17)}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+5}+\dfrac1{x+5}-\dfrac1{x+10}+\dfrac1{x+10}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac{15}{(x+2)(x+17)}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow x=15.$

b)

Điều kiện: $x\ne1,3,8,20.$

$\dfrac2{(x-1)(x-3)}+\dfrac5{(x-3)(x-8)}+\dfrac{12}{(x-8)(x-20)}-\dfrac1{x-20}=-\dfrac34$

$\Leftrightarrow\left(\dfrac1{x-3}-\dfrac1{x-1}\right)+\left(\dfrac1{x-8}-\dfrac1{x-3}\right)+\left(\dfrac1{x-20}-\dfrac1{x-8}\right)-\dfrac1{x-20}=-\dfrac34$

$\Leftrightarrow-\dfrac1{x-1}=-\dfrac34$

$\Leftrightarrow\dfrac1{x-1}=\dfrac34$

$\Leftrightarrow4=3(x-1)$

$\Leftrightarrow3x=7$

$\Leftrightarrow x=\dfrac73.$