So sánh 18/49 và 6/37
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Bài 1 :
1) Ta có : \(\frac{19}{18}=1+\frac{1}{18}\)
\(\frac{2019}{2018}=1+\frac{1}{2018}\)
Vì \(\frac{1}{18}>\frac{1}{2018}\)
Nên : \(\frac{19}{18}>\frac{2019}{2018}\)
Ta có:
\(\dfrac{37}{-49}< 0;\dfrac{-12}{-35}=\dfrac{12}{35}>0\)
\(\Rightarrow\dfrac{37}{-49}< \dfrac{-12}{-35}\)
Vậy...
\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(=\frac{17.2}{7.13}+\frac{17.3}{13.22}+\frac{17.5}{22.37}+\frac{17.4}{37.49}\)
\(=17\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+...+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)\(=\frac{17}{3}.\frac{6}{49}=\frac{17.2}{49}=\frac{34}{49}\)
Có : \(\frac{17}{24}=\frac{34}{48}\)
Vì 48 < 49 => \(\frac{34}{48}>\frac{34}{49}\). Hay \(\frac{17}{24}>A\)
a: -15/37>-25/37
b: -13/21=-26/42
-9/14=-27/42
mà -26>-42
nên -13/21>-9/14
c: -49/-63=7/9
56/80=7/10
=>-49/-63>56/80
d: 3/14=1-11/14
4/15=1-11/15
mà 11/14>11/15
nên 3/14<4/15
\(\dfrac{3}{4}< \dfrac{3+3}{4+3}=\dfrac{6}{7}\)
\(\dfrac{6}{7}< \dfrac{6+3}{7+3}< \dfrac{9}{10}\)
\(\dfrac{9}{10}< \dfrac{9+9}{10+9}=\dfrac{18}{19}\)
\(\dfrac{18}{19}< \dfrac{18+18}{19+18}=\dfrac{36}{37}\)
Vậy \(\dfrac{36}{37}>\dfrac{18}{19}>\dfrac{9}{10}>\dfrac{6}{7}>\dfrac{3}{4}\)
\(\dfrac{18}{19}=1-\dfrac{1}{19}\)
\(\dfrac{9}{10}=1-\dfrac{1}{10}\)
\(\dfrac{6}{7}=1-\dfrac{1}{7}\)
\(\dfrac{3}{4}=1-\dfrac{1}{4}\)
\(\dfrac{36}{37}=1-\dfrac{1}{37}\)
Do \(37>19>10>7>4\)
\(\Rightarrow\dfrac{1}{37}< \dfrac{1}{19}< \dfrac{1}{10}< \dfrac{1}{7}< \dfrac{1}{4}\)
\(\Rightarrow1-\dfrac{1}{37}>1-\dfrac{1}{19}>1-\dfrac{1}{10}>1-\dfrac{1}{7}>1-\dfrac{1}{4}\)
\(\Rightarrow\dfrac{36}{37}>\dfrac{18}{19}>\dfrac{9}{10}>\dfrac{6}{7}>\dfrac{3}{4}\)
a: 27^4=(3^3)^4=3^12<3^18
b: 49^4=(7^2)^4=7^8
c: 9^16=(9^2)^8=81^8>27^8
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