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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: \(=\dfrac{13}{19}\cdot\dfrac{19}{26}-\dfrac{13}{19}\cdot\dfrac{71}{43}+\dfrac{71}{43}\cdot\dfrac{13}{19}-\dfrac{71}{43}\cdot\dfrac{86}{71}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
a,
\(\dfrac{13}{17}=1-\dfrac{4}{17}\\ \dfrac{25}{29}=1-\dfrac{4}{29}\\ \dfrac{4}{17}>\dfrac{4}{29}\Rightarrow1-\dfrac{4}{17}< 1-\dfrac{4}{29}\Leftrightarrow\dfrac{13}{17}< \dfrac{25}{29}\)
Vậy \(\dfrac{13}{17}< \dfrac{25}{29}\)
b,
\(\dfrac{59}{101}>\dfrac{56}{101}>\dfrac{56}{105}\\ \Rightarrow\dfrac{59}{101}>\dfrac{56}{105}\)
Vậy \(\dfrac{59}{101}>\dfrac{56}{105}\)
c,
\(\dfrac{14}{55}>\dfrac{14}{56}=\dfrac{1}{4}=\dfrac{20}{80}>\dfrac{20}{83}\)
Vậy \(\dfrac{14}{55}>\dfrac{20}{83}\)
1: \(=\dfrac{15}{37}\cdot\dfrac{38}{41}-\dfrac{15}{37}\cdot\dfrac{74}{45}-\dfrac{38}{41}\cdot\dfrac{15}{37}-\dfrac{38}{41}\cdot\dfrac{82}{76}\)
\(=\dfrac{-2}{3}-1=-\dfrac{5}{3}\)
2: \(=\dfrac{47}{53}\cdot\dfrac{17}{3}-\dfrac{47}{53}\cdot\dfrac{53}{47}+\dfrac{17}{3}\cdot\dfrac{6}{17}-\dfrac{17}{3}\cdot\dfrac{47}{53}\)
\(=-1+2=1\)
b, Ta có: \(\dfrac{58}{53}>1>\dfrac{36}{55}\)
hay \(\dfrac{58}{53}>\dfrac{36}{55}\)
\(\Rightarrow0-\dfrac{58}{53}< 0-\dfrac{36}{55}\)
\(\Rightarrow\dfrac{-58}{53}< \dfrac{-36}{55}\)
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: \(\dfrac{-1941}{1931}< \dfrac{-2011}{2001}\)
Lời giải:
a, \(\dfrac{12}{19}.\dfrac{7}{15}.\dfrac{-13}{17}.\dfrac{19}{12}.\dfrac{17}{13}\)
\(=\dfrac{12}{19}.\dfrac{-7}{15}.\dfrac{13}{17}.\dfrac{19}{12}.\dfrac{17}{13}\)
\(=\left(\dfrac{12}{19}.\dfrac{19}{12}\right).\left(\dfrac{13}{17}.\dfrac{17}{13}\right).\dfrac{-7}{15}\)
\(=1.1.\dfrac{-7}{15}\)
\(=\dfrac{-7}{15}\)
b, \(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{27}=\dfrac{-5}{9}\\\dfrac{-45}{b}=\dfrac{-5}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}9a=-5.27\\-5b=-45.9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}9a=-135\\-5b=-405\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=-15\\b=-81\end{matrix}\right.\)
Vậy, \(a=-15;b=-81\).
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