1: \(\dfrac{4}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)

\(=1+\dfrac{1}{2}\)

\(=\dfrac{3}{2}\)

2: \(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)

\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)

\(=\dfrac{1}{3}\)

\(\frac{2^{10}.9^{41}.25^{23}}{3^{50}.15^{35}.10^3}\)=\(\frac{2^{10}.\left(3^3\right)^{41}.\left(5^2\right)^{23}}{3^{50}.\left(3.5\right)^{35}.\left(2.5\right)^3}\)=\(\frac{2^{10}.3^{123}.5^{46}}{3^{50}.3^{35}.5^{35}.2^3.5^3}\)=\(\frac{2^{10}.3^{123}.5^{46}}{3^{85}.5^{38}.2^3}\)=\(2^7.3^{38}.5^8\)

a: Ta có: \(\left(-\frac57-\frac79+\frac{9}{11}-\frac{11}{13}\right)\cdot\left(3-\frac34\right)\)

\(=\left(-\frac{6435}{9009}-\frac{7007}{9009}+\frac{7371}{9009}-\frac{7623}{9009}\right)\cdot\frac94\)

\(=-\frac{13694}{9009}\cdot\frac94=\frac{-6847}{2}\cdot\frac{1}{1001}=\frac{-6847}{2002}\)

b: \(\left(\frac{10}{21}+\frac{14}{27}-\frac{6}{11}+\frac{22}{39}\right):\left(2-\frac23\right)\)

\(=\left(\frac{12870}{27027}+\frac{14014}{27027}-\frac{14742}{27027}+\frac{15246}{27027}\right):\frac43\)

\(=\frac{27388}{27027}\cdot\frac34=\frac{6847}{9009}\)