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Lời giải:
$\frac{179}{197}< 1< \frac{971}{917}$
$\frac{183}{184}> 0> \frac{-184}{183}$
a) 179/197 là số bé
971/917 là số lớn
=>179/197<971/917
b)183/184 là số bé
-184/-183=184/183 là số lớn
=> 183/184<184/183
a) Do 179/197 < 1
971/917 > 1
=> 179/197 < 971/917
b) Do 183/184 < 1
-184/-183 = 184/183 > 1
=> 183/184 < -184/-183
1/Vì 179/197<1 ; 971/917>1
=>179/197<971/917
2/Vì 183/184<1 ; 184/183>1
=>183/184<184/183
3/Ta có : -3/31=-3*101/31*101=-303/3131
Vì -303>-789 =>-303/3131>-789/3131 =>-3/31>-789/3131
1) Ta có: \(\frac{179}{197}<1;\frac{971}{917}>1\)
=> \(\frac{179}{197}<1<\frac{971}{917}\)
=> \(\frac{179}{197}<\frac{971}{917}\)
2) Ta có: \(\frac{183}{184}<1;\frac{184}{183}>1\)
=> \(\frac{183}{184}<1<\frac{184}{183}\)
=> \(\frac{183}{184}<\frac{184}{183}\)
Câu 1:
1) 179/197 và 971/917
Ta có:
\(1-\frac{179}{197}=\frac{18}{197}\)
\(1-\frac{971}{917}=\frac{-54}{917}\)
Mà \(\frac{-54}{917}<\frac{18}{197}\)
\(\Rightarrow\frac{971}{917}<\frac{179}{197}\)
Câu 2:
Ta có:
\(1-\frac{183}{184}=\frac{1}{184}\)
\(1-\frac{184}{183}=\frac{-1}{183}\)
Mà:\(\frac{-1}{183}<\frac{1}{184}\)
\(\Rightarrow\frac{184}{183}<\frac{183}{184}\)
\(\dfrac{-178}{179}>-1>\dfrac{-191}{189}\\ \dfrac{127}{129}=1-\dfrac{2}{129};\dfrac{871}{873}=1-\dfrac{2}{873}\\ \dfrac{2}{129}>\dfrac{2}{873}\left(129< 873\right)\Leftrightarrow1-\dfrac{2}{129}< 1-\dfrac{2}{873}\Leftrightarrow\dfrac{127}{129}< \dfrac{871}{873}\)
Ta có:
\(A=\dfrac{196}{197}+\dfrac{197}{198}\)
\(B=\dfrac{196+197}{197+198}\)
\(=\dfrac{196}{197+198}+\dfrac{197}{197+198}\)
Áp dụng tính chất \(\dfrac{a}{b}>\dfrac{a}{b+m}\) ta có:
\(\left\{{}\begin{matrix}\dfrac{196}{197}>\dfrac{196}{197+198}\\\dfrac{197}{198}>\dfrac{197}{197+198}\end{matrix}\right.\)
\(\Rightarrow\dfrac{196}{197}+\dfrac{197}{198}>\dfrac{196}{197+198}+\dfrac{197}{197+198}=\dfrac{196+197}{197+198}\)
Vậy \(A>B\)
2/
a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)
\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)
-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)
b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)
-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)
c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)
\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)
-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
a) \(\dfrac{23}{24}< 1\)
\(\dfrac{24}{23}>1\)
\(\Rightarrow\dfrac{23}{24}< \dfrac{24}{23}\)
b) \(\dfrac{4}{21}< \dfrac{4}{20}=\dfrac{1}{5}=\dfrac{6}{30}< \dfrac{6}{29}\)
c) \(\dfrac{6}{7}=1-\dfrac{1}{7}< \dfrac{8}{9}=1-\dfrac{1}{9}\)
d) \(\dfrac{1212}{1313}=\dfrac{12\times101}{13\times101}=\dfrac{12}{13}\)
d)
Ta có: \(\dfrac{1}{51}>\dfrac{1}{100}\)
\(\dfrac{1}{52}>\dfrac{1}{100}\)
...
\(\dfrac{1}{99}>\dfrac{1}{100}\)
\(\dfrac{1}{100}=\dfrac{1}{100}\)
\(\Rightarrow S=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{99}+\dfrac{1}{100}>\dfrac{1}{100}.50=\dfrac{1}{2}\)\(\Rightarrow S>\dfrac{1}{2}\)
các con trên ???