so sánh qua số trung gian:
a)\(\dfrac{12}{49}\);\(\dfrac{13}{47}\)
b)\(\dfrac{12}{47}\);\(\dfrac{19}{77}\)
giúp mk với
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Ta có:
\(\dfrac{37}{-49}< 0;\dfrac{-12}{-35}=\dfrac{12}{35}>0\)
\(\Rightarrow\dfrac{37}{-49}< \dfrac{-12}{-35}\)
Vậy...
Ta có:
\(\frac{25}{12}>\frac{24}{12}=2\)
\(\frac{49}{25}< \frac{50}{25}=2\)
Vì \(\frac{25}{12}>2>\frac{49}{25}\) nên \(\frac{25}{12}>\frac{49}{25}\)
Chúc bạn học tốt!
Ta có:
\(\dfrac{-49}{211}< 0;\dfrac{13}{1999}>0\)
⇒ \(\dfrac{-49}{211}< \dfrac{13}{1999}\)
a) Ta có \(\dfrac{23}{27}>\dfrac{23}{29};\dfrac{23}{29}>\dfrac{22}{29}\)
Vậy \(\dfrac{23}{27}>\dfrac{22}{29}\)
b) Ta có \(\dfrac{15}{25}=1-\dfrac{2}{5}\)
\(\dfrac{25}{49}=1-\dfrac{24}{49}\)
Vì \(\dfrac{2}{5}=\dfrac{24}{60}< \dfrac{24}{49}\)
Vậy \(\dfrac{15}{25}>\dfrac{25}{49}\)
\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)
\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)
\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)
Áp dụng t/c dtsbn:
\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)
\(\dfrac{-11}{-32}>\dfrac{16}{49}\)
\(\dfrac{-2020}{-2021}>\dfrac{-2021}{2022}\)
a: \(\dfrac{12}{49}< \dfrac{13}{49}< \dfrac{13}{47}\)
b: \(\dfrac{12}{47}>\dfrac{19}{47}>\dfrac{19}{77}\)
a/ so sánh bằng 12/47
b/ chưa tìm ra