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\(a,\dfrac{1}{7}+\dfrac{1}{3}=\dfrac{3}{21}+\dfrac{7}{21}=\dfrac{10}{21}\\ b,\dfrac{-2}{3}+\dfrac{-5}{7}=\dfrac{-14+\left(-15\right)}{21}=\dfrac{-29}{21}\\ c,\dfrac{1}{2}-\dfrac{-1}{2}-\dfrac{1}{3}=\dfrac{3+3-2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(a.\dfrac{4}{28}+\dfrac{12}{36}=\dfrac{1}{7}+\dfrac{1}{3}=\dfrac{3}{21}+\dfrac{7}{21}=\dfrac{10}{21}\\ b.\dfrac{-12}{18}+\dfrac{-15}{21}=\dfrac{-2}{3}+\dfrac{-5}{7}=\dfrac{-14}{21}+\dfrac{-15}{21}=\dfrac{-29}{21}\\ c.\dfrac{14}{28}+\dfrac{16}{32}-\dfrac{17}{51}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{17}{51}=1-\dfrac{17}{51}=\dfrac{2}{3}\)
\(\dfrac{\left(17\dfrac{8}{19}-16\dfrac{9}{18}\right).\left(17,5+16\dfrac{17}{51}-32\dfrac{15}{22}\right)}{\dfrac{7}{3.13}+\dfrac{7}{13.23}+\dfrac{7}{23.33}}\)
=\(\dfrac{\dfrac{35}{38}.\dfrac{38}{33}}{\dfrac{7}{10}\left(\dfrac{1}{3}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{23}+\dfrac{1}{23}-\dfrac{1}{33}\right)}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{10}.\left(\dfrac{1}{3}-\dfrac{1}{33}\right)}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{10}.\dfrac{10}{33}}\)
=\(\dfrac{\dfrac{35}{33}}{\dfrac{7}{33}}\)
=\(\dfrac{35}{33}:\dfrac{7}{33}\)
=\(\dfrac{35}{33}.\dfrac{33}{7}\)
=5
9: \(=\dfrac{47}{51}\cdot\dfrac{17}{94}-\dfrac{47}{51}\cdot\dfrac{53}{91}-\dfrac{53}{91}\cdot\dfrac{91}{53}+\dfrac{53}{91}\cdot\dfrac{47}{51}\)
\(=\dfrac{1}{6}-1=-\dfrac{5}{6}\)
10: \(=\dfrac{13}{19}\cdot\dfrac{19}{26}-\dfrac{13}{19}\cdot\dfrac{71}{43}+\dfrac{71}{43}\cdot\dfrac{13}{19}-\dfrac{71}{43}\cdot\dfrac{86}{71}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)
\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)
\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)
b, \(K =\) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}\)
\(K = \) \(\dfrac{3}{4}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}\)
\(K = \) \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{18}{21}+\dfrac{3}{21}\right)+\left(\dfrac{19}{32}+\dfrac{13}{32}\right)\)
\(K = \) \(1 + 1 + 1\)
\(K = \) \(3\)
29/19.49/51+29/19.34/51-29/19.32/51
=29/19.(49/51+34/51-32/51)
=29/19.1
=29/19
\(\dfrac{29}{19}.\dfrac{49}{51}+\dfrac{29}{19}.\dfrac{34}{51}-\dfrac{29}{19}.\dfrac{32}{51}\)
\(=\dfrac{29}{19}.\left(\dfrac{49}{51}+\dfrac{34}{51}-\dfrac{32}{51}\right)\)
\(=\dfrac{29}{19}.1\)
\(=\dfrac{29}{19}\)
a: \(=\dfrac{-12}{7}\left(\dfrac{4}{35}+\dfrac{31}{35}\right)-\dfrac{2}{7}=\dfrac{-12}{7}-\dfrac{2}{7}=-2\)
b: =(-4)+(-4)+...+(-4)
=-4*25=-100
c: \(=157\cdot\left(-37\right)-41\cdot53+37\cdot157+51\cdot53\)
=10*53
=530
1. \(-\dfrac{52}{17}+\left(\dfrac{12}{19}+\dfrac{52}{17}\right)=\left[\left(-\dfrac{52}{17}\right)+\left(-\dfrac{52}{17}\right)\right]+\dfrac{12}{19}=\dfrac{12}{19}\)
2. \(\dfrac{21}{35}+\left(-1+\dfrac{14}{35}\right)=\dfrac{3}{5}+\left(-1+\dfrac{2}{5}\right)=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(-1\right)=1-1=0\)
1. \(\frac{-52}{17}+\left(\frac{12}{19}+\frac{52}{17}\right)=\frac{-52}{17}+\frac{12}{19}+\frac{52}{17}\)\(=\left(\frac{-52}{17}+\frac{52}{17}\right)+\frac{12}{19}=\frac{0}{17}+\frac{12}{19}=\frac{12}{19}\)
2. \(\frac{21}{35}+\left(-1+\frac{14}{35}\right)=\frac{21}{35}-1+\frac{14}{35}=\left(\frac{21}{35}+\frac{14}{35}\right)-1=1-1=0\)
\(=-\left[\dfrac{-17}{21}+\dfrac{-19}{51}-\dfrac{4}{21}\right]+\dfrac{32}{51}\)
\(=1+\dfrac{19}{51}+\dfrac{32}{51}=1+1=2\)