$\textbf{a)}$

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

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

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

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

$\textbf{b)}$

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

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

$=\dfrac19.$

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

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

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

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

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

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

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

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 ~

Biểu thức 1

$A=3\dfrac12\cdot\dfrac4{49}\cdot\left(2,(4)\cdot2\dfrac5{11}\right):\left(-\dfrac{42}{5}\right)$

$=\dfrac72\cdot\dfrac4{49}\cdot\left(\dfrac{22}{9}\cdot\dfrac{27}{11}\right)\cdot\dfrac5{-42}$

$=\dfrac72\cdot\dfrac4{49}\cdot6\cdot\dfrac5{-42}$

$=\dfrac{12}{7}\cdot\dfrac5{-42}$

$=\dfrac{60}{-294}$

$=-\dfrac{10}{49}.$

Biểu thức 2

$B=\left[0,(5)\cdot0,(2)\right]:\left(3\dfrac13:\dfrac{33}{25}\right)-\left(\dfrac25\cdot1\dfrac13\right):\dfrac43$

$=\left(\dfrac59\cdot\dfrac29\right):\left(\dfrac{10}{3}:\dfrac{33}{25}\right)-\left(\dfrac25\cdot\dfrac43\right):\dfrac43$

$=\dfrac{10}{81}:\dfrac{250}{99}-\dfrac{8}{15}\cdot\dfrac34$

$=\dfrac{10}{81}\cdot\dfrac{99}{250}-\dfrac25$

$=\dfrac{11}{225}-\dfrac{90}{225}$

$=-\dfrac{79}{225}.$

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

$\textbf{A)}$

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

$\text{Vì }1000-10^3=1000-1000=0.$

$\Rightarrow A=0.$

c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)

a.211,0465116

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