$B=\left(\dfrac1{2^2}-1\right)\left(\dfrac1{3^2}-1\right)\cdots\left(\dfrac1{99^2}-1\right)$

$=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)\cdots\left(-\dfrac{99^2-1}{99^2}\right)$

$=(-1)^{98}\prod_{k=2}^{99}\dfrac{(k-1)(k+1)}{k^2}$

$=\left(\prod_{k=2}^{99}\dfrac{k-1}{k}\right)\left(\prod_{k=2}^{99}\dfrac{k+1}{k}\right)$

$=\left(\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{98}{99}\right)\left(\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{100}{99}\right)$

$=\dfrac1{99}\cdot\dfrac{100}{2}$

$=\dfrac{50}{99}.$

Những câu từ D trở đi là các câu riêng biệt ak bạn

\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)

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

$\textbf{A)}$

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

$\text{Vì }1000-10^3=1000-1000=0.$

$\Rightarrow A=0.$

$\textbf{a)}$

$\left(-\dfrac34+\dfrac27\right):\dfrac27+\left(-\dfrac14+\dfrac57\right):\dfrac23$

$=\left(-\dfrac{13}{28}\right)\cdot\dfrac72+\dfrac{13}{28}\cdot\dfrac32$

$=-\dfrac{13}{8}+\dfrac{39}{56}$

$=-\dfrac{13}{14}.$

$\textbf{b)}$

$\left(-\dfrac13\right)^2\cdot\dfrac4{11}+\dfrac7{11}\cdot\left(-\dfrac13\right)^2$

$=\dfrac19\left(\dfrac4{11}+\dfrac7{11}\right)$

$=\dfrac19.$

a) \(\frac{\left(-1\right)}{4}^2+\frac{3}{8}.\left(\frac{-1}{6}\right)-\frac{3}{16}:\left(\frac{-1}{2}\right)=\left(\frac{-1}{4}\right)^2+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\left(\frac{1}{16}\right)+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\frac{5}{272}-\left(\frac{-3}{8}\right)=\frac{107}{272}\)

$\textbf{a)}$

$A=\left(-\dfrac14\right)^2+\dfrac38\cdot\left(-\dfrac16\right)-\dfrac3{16}:\left(-\dfrac12\right)$

$=\dfrac1{16}-\dfrac1{16}+\dfrac38$

$=\dfrac38.$

\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)

\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)

\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)

\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)

\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)

\(=\left[\frac{2}{3}-\left(-1\right)+1\right]:\frac{-4}{3}\)

\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)

~ Hok tốt ~

$A=\dfrac{\left(\dfrac49\right)^2\cdot\left(-\dfrac9{16}\right)\cdot(-1)^{19}}{\left(\dfrac4{25}\right)^2\cdot\left(-\dfrac{25}{144}\right)^2\cdot\left(-\dfrac{49}{144}\right)^2}$

$=\dfrac{\dfrac{16}{81}\cdot\left(-\dfrac9{16}\right)\cdot(-1)}{\dfrac{16}{625}\cdot\dfrac{625}{144^2}\cdot\dfrac{49^2}{144^2}}$

$=\dfrac{\dfrac19}{\dfrac{49^2}{144^4}}$

$=\dfrac19\cdot\dfrac{144^4}{49^2}$

$=\dfrac19\cdot\dfrac{(2^4\cdot3^2)^4}{7^4}$

$=\dfrac19\cdot\dfrac{2^{16}\cdot3^8}{7^4}$

$=\dfrac{2^{16}\cdot3^6}{7^4}$

$=\dfrac{65536\cdot729}{2401}$

$=\dfrac{47775744}{2401}.$