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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. \(\frac{21}{35}+\left(-1+\frac{14}{35}\right)=\frac{21}{35}-1+\frac{14}{35}=\left(\frac{21}{35}+\frac{14}{35}\right)-1=1-1=0\)
a)\(12< 13;49>47\)
\(\Rightarrow\dfrac{12}{49}< \dfrac{13}{47}\)
b)\(\dfrac{64}{85}>\dfrac{43}{85}\Rightarrow\dfrac{64}{85}>\dfrac{1}{2}\)
\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{17}{35}< \dfrac{64}{85}\)
c) \(\dfrac{19}{31}>\dfrac{16}{31}\Rightarrow\dfrac{19}{31}>\dfrac{1}{2}\)
\(\dfrac{17}{35}< \dfrac{17}{34}\Rightarrow\dfrac{17}{35}< \dfrac{1}{2}\)
\(\Rightarrow\dfrac{17}{35}< \dfrac{19}{31}\)
d)
\(1-\dfrac{67}{77}=\dfrac{10}{77}\)
\(1-\dfrac{73}{83}=\dfrac{10}{83}\)
\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)
e)\(1-\dfrac{456}{461}=\dfrac{5}{461}\)
\(1-\dfrac{123}{128}=\dfrac{5}{128}\)
\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
\(a,\dfrac{12}{49}< \dfrac{12}{47}< \dfrac{13}{47}\Rightarrow\dfrac{12}{49}< \dfrac{12}{47}\)
b, Ta có: \(\dfrac{17}{35}=\dfrac{51}{105}\)
\(\dfrac{64}{85}>\dfrac{64}{105}>\dfrac{51}{105}\Rightarrow\dfrac{64}{85}>\dfrac{51}{105}\) hay \(\dfrac{64}{85}>\dfrac{17}{85}\)
c,\(\dfrac{19}{31}>\dfrac{17}{31}>\dfrac{17}{35}\Rightarrow\dfrac{19}{31}>\dfrac{17}{35}\)
d, \(\dfrac{67}{77}+\dfrac{10}{77}=1\)
\(\dfrac{73}{83}+\dfrac{10}{83}=1\)
\(\dfrac{10}{77}>\dfrac{10}{83}\Rightarrow\dfrac{67}{77}< \dfrac{73}{83}\)
e, \(\dfrac{456}{461}+\dfrac{5}{461}=1\)
\(\dfrac{123}{128}+\dfrac{5}{128}=1\)
\(\dfrac{5}{461}< \dfrac{5}{128}\Rightarrow\dfrac{456}{461}>\dfrac{123}{128}\)
b, \(K =\) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}\)
\(K = \) \(\dfrac{3}{4}+\dfrac{18}{21}+\dfrac{19}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}\)
\(K = \) \(\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{18}{21}+\dfrac{3}{21}\right)+\left(\dfrac{19}{32}+\dfrac{13}{32}\right)\)
\(K = \) \(1 + 1 + 1\)
\(K = \) \(3\)
7)\(\dfrac{-19}{34}\left(\dfrac{17}{19}+\dfrac{49}{18}\right)+\dfrac{49}{18}\left(\dfrac{19}{34}-\dfrac{18}{7}\right)\)
=\(\dfrac{-19}{34}.\dfrac{17}{19}+\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}-\dfrac{18}{7}.\dfrac{49}{18}\)
=\(\dfrac{1}{2}+\left(\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}\right)-7\)
=\(\dfrac{1}{2}+\left[\dfrac{49}{18}\left(\dfrac{-19}{34}+\dfrac{19}{34}\right)\right]-7\)
=\(\dfrac{1}{2}+0-7=\dfrac{-13}{2}\)
8)\(\dfrac{29}{32}\left(\dfrac{41}{36}-\dfrac{32}{58}\right)-\dfrac{41}{36}\left(\dfrac{29}{32}+\dfrac{18}{41}\right)\)
=\(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{29}{32}.\dfrac{32}{58}-\dfrac{41}{36}.\dfrac{29}{32}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(\left(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{41}{36}\dfrac{29}{32}\right)-\dfrac{29}{32}.\dfrac{32}{58}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(0-\dfrac{1}{2}+\dfrac{1}{2}=0\)
Câu 1:
a) \(\dfrac{-15}{17}\) và \(\dfrac{-19}{21}\)
Ta có: \(\dfrac{-15}{17}=-1+\dfrac{2}{17}\); \(\dfrac{-19}{21}=-1+\dfrac{2}{21}\)
Vì \(\dfrac{2}{17}>\dfrac{2}{21}\)
Do đó: \(\dfrac{-15}{17}>\dfrac{19}{-23}\)
b) \(\dfrac{-13}{19}\) và \(\dfrac{19}{-23}\)
Ta có: \(\dfrac{19}{23}>\dfrac{19}{25}\); \(\dfrac{13}{19}=1-\dfrac{6}{19}\); \(\dfrac{19}{25}=1-\dfrac{6}{25}\)
mà \(\dfrac{6}{19}>\dfrac{6}{25}\) \(\Rightarrow\dfrac{13}{19}< \dfrac{19}{25}< \dfrac{19}{23}\)
Vì \(\dfrac{13}{19}< \dfrac{19}{23}\Rightarrow\dfrac{-13}{19}>\dfrac{19}{-23}\)
c) \(\dfrac{-24}{35}\) và \(\dfrac{-19}{30}\)
Ta có: \(\dfrac{-24}{35}=-1+\dfrac{19}{35}\);\(\dfrac{-19}{30}=-1+\dfrac{11}{30}\)
Vì \(\dfrac{11}{35}< \dfrac{11}{30}\)
Do đó: \(\dfrac{-24}{35}< \dfrac{-19}{30}\)
d) \(\dfrac{-1941}{1931}\) và \(\dfrac{-2011}{2001}\); \(\dfrac{-2011}{2001}=-1+\dfrac{10}{2001}\)
Vì \(\dfrac{10}{1931}< \dfrac{10}{1001}\)
Do đó: \(\dfrac{-1941}{1931}< \dfrac{-2011}{2001}\)
Ta có: \(\dfrac{-1941}{1931}=-1+\dfrac{10}{1931}\)
Sorry câu d mình viết ngược:
Làm lại:
d) \(\dfrac{-1941}{1931}\) và \(\dfrac{-2011}{2001}\)
Ta có: \(\dfrac{-1941}{1931}=-1+\dfrac{10}{1931};\)
\(\dfrac{-2011}{2001}=-1+\dfrac{10}{2001}\)
Vì \(\dfrac{10}{1931}< \dfrac{10}{1001}\)
Do đó: \(\dfrac{-1941}{1931}< \dfrac{-2011}{2001}\)
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
\(D=\dfrac{1}{2}+\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}+\dfrac{-3}{35}+\dfrac{1}{3}+\dfrac{1}{41}\)
\(D=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{-3}{35}\right)+\dfrac{1}{41}\)
\(D=1+-1+\dfrac{1}{41}\)
\(D=0+\dfrac{1}{41}\)
\(D=\dfrac{1}{41}\)
\(C=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}+\dfrac{-1}{36}+\dfrac{-2}{9}\right)+\dfrac{1}{57}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-1-8}{36}+\dfrac{1}{57}\)
=1/57
\(E=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}=\dfrac{1}{127}\)
\(P=\dfrac{32}{64}\cdot\dfrac{-57}{19}+\dfrac{35}{21}\cdot\dfrac{22}{44}=\dfrac{1}{2}\left(-3+\dfrac{5}{3}\right)=\dfrac{1}{2}\cdot\dfrac{-4}{3}=\dfrac{-2}{3}\)
\(Q=\dfrac{75}{125}\cdot\dfrac{82}{164}+\dfrac{49}{98}\cdot\dfrac{-35}{105}=\dfrac{1}{2}\left(\dfrac{3}{5}-\dfrac{1}{3}\right)=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)