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`17/20 = 119/140`
`11/14=110/140`
`=> 119/140 > 110/140`
`=> 17/20 > 11/14`
\(\dfrac{17}{20}=\dfrac{17\cdot14}{20\cdot14}=\dfrac{238}{280}\)
\(\dfrac{11}{14}=\dfrac{11\cdot20}{14\cdot20}=\dfrac{220}{280}\)
mà 238>220
nên \(\dfrac{17}{20}>\dfrac{11}{14}\)
Ta có:11/12 và 17/18
⇒11x18<12x17 (198<204)
CHÚC BẠN HỌC TỐT!!!!
a) \(A=2A-A\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1-\dfrac{1}{2^{2022}}\)
b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)
\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)
Lời giải:
\(A=1.3.5.7...99=\frac{1.2.3.4...99.100}{2.4.6.8.100}=\frac{1.2.3...99.100}{(1.2)(2.2)(3.2)...(50.2)}\)
\(=\frac{1.2.3...99.100}{(1.2.3...50).2^{50}}=\frac{51.52...100}{2^{50}}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}=B\)
\(\dfrac{-7}{-17}=\dfrac{7}{17}\)
Vì 7>6 nên \(\dfrac{-7}{-17}>\dfrac{6}{17}\)
\(\dfrac{-7}{-17}=\dfrac{7}{17}\)
\(\dfrac{6}{17}\) giữ nguyên
Vì \(7>6\)
\(\Rightarrow\dfrac{-7}{-17}>\dfrac{6}{11}\)
a: 43/52>26/52=1/2=60/120
b: 17/68=1/4<1/3=35/105<35/103
c: \(\dfrac{2018\cdot2019-1}{2018\cdot2019}=1-\dfrac{1}{2018\cdot2019}\)
\(\dfrac{2019\cdot2020-1}{2019\cdot2020}=1-\dfrac{1}{2019\cdot2020}\)
2018*2019<2019*2020
=>-1/2018*2019<-1/2019*2020
=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
Giải:
Ta dùng tích chéo để giải.
Ta thấy:
\(11.60=660.\)
\(17.52=884.\)
mà \(660< 884.\)
\(\Rightarrow\dfrac{11}{52}< \dfrac{17}{60}.\)
Vậy...
~ Học tốt!!! ~
Ta kiểm tra hai tích chéo:
Ta có: 11.60 = 660
17.52 = 884
Vì 660 < 884 nên \(\dfrac{11}{52}< \dfrac{17}{60}\)