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ta có: \(1-\frac{17}{20}=\frac{3}{20};1-\frac{22}{25}=\frac{3}{25}\)
\(\Rightarrow\frac{3}{20}>\frac{3}{25}\Rightarrow1-\frac{17}{20}>1-\frac{22}{25}\)
\(\Rightarrow\frac{17}{20}< \frac{22}{25}\)
Ta có : \(\frac{11}{12}=\frac{11.10}{12.10}=\frac{110}{120}\)
\(\frac{9}{10}=\frac{9.12}{10.12}=\frac{108}{120}\)
Ta thấy \(\frac{110}{120}>\frac{108}{120}\Rightarrow\frac{11}{20}>\frac{9}{10}\)
Vậy \(\frac{11}{20}>\frac{9}{10}\)
Cần nhớ:
\(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+n}{b+n}\left(n\inℕ^∗\right)\)
\(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{b+n}\left(n\inℕ^∗\right)\)
Ta thấy:
\(\frac{19}{29}< 1\Rightarrow\frac{19}{29}< \frac{19+1}{29+1}=\frac{20}{30}=\frac{2}{3}\)
Ta lại có:
\(\frac{2}{3}=\frac{2.7}{3.7}=\frac{14}{21}< 1\Rightarrow\frac{14}{21}< \frac{14+6}{21+6}=\frac{20}{27}=\frac{20.3}{27.3}=\frac{60}{81}\)
\(\Rightarrow\frac{19}{29}< \frac{60}{81}\) (1)
Ta có:
\(\frac{60}{81}=\frac{20}{27}< 1\Rightarrow\frac{20}{27}< \frac{20+1}{27+1}=\frac{21}{28}< \frac{21}{25}\)
=>\(\frac{60}{81}< \frac{21}{25}\) (2)
Từ (1) và (2) => \(\frac{19}{29}< \frac{60}{81}< \frac{21}{25}\)
\(\frac{21}{25}=\frac{49329}{58725}\)
\(\frac{60}{81}=\frac{43500}{58725}\)
\(\frac{19}{29}=\frac{38475}{58725}\)
\(\)vì \(\frac{49329}{58725}>\frac{43500}{58725}>\frac{38475}{58725}\)nên\(\frac{21}{25}>\frac{60}{81}>\frac{19}{29}\)
\(\frac{21}{25};\frac{60}{81};\frac{19}{29}\)
Ta có : \(\frac{60}{81}=\frac{60:3}{81:3}=\frac{20}{27}\); giữ nguyên \(\frac{21}{25};\frac{19}{29}\).
Vì \(\frac{21}{25}>\frac{20}{25}>\frac{20}{27}\)nên \(\frac{21}{25}>\frac{20}{27}\)
Vì \(\frac{20}{27}>\frac{20}{29}>\frac{19}{29}\)nên \(\frac{20}{27}>\frac{19}{29}\). Vậy :
\(\frac{21}{25}>\frac{20}{27}>\frac{19}{29}\)hay \(\frac{21}{25}>\frac{60}{81}>\frac{19}{29}\)
\(\frac{18}{75}=\frac{6}{25}\)
\(\frac{28}{112}=\frac{1}{4}=\frac{6}{24}\)
Vì 25>24 nên \(\frac{6}{25}< \frac{6}{24}\Leftrightarrow\frac{18}{75}>\frac{28}{112}\)
So sánh: \(\frac{23}{48}< \frac{47}{92}\)(Nhân chéo tử này với mẫu kia bên nào có kết quả lớn hơn thì bên đó lớn hơn bạn nhekk)
Ta có \(\frac{23}{48}< \frac{23}{46}=\frac{46}{92}< \frac{47}{92}\)
Vậy \(\frac{23}{48}< \frac{47}{92}\)
a; (5142 - 17 x 8 + 242 : 11) x (27 - 3 x 9)
= (5142 - 17 x 8 + 242 : 11) x (27 - 27)
= (5142 - 17 x 8 + 242 : 11) x 0
= 0
b;
(1 + \(\dfrac{1}{2}\)) \(\times\) (1 + \(\dfrac{1}{3}\)) \(\times\) ( 1 + \(\dfrac{1}{4}\)) \(\times\) ... \(\times\) (1 + \(\dfrac{1}{2010}\)) \(\times\)(1 + \(\dfrac{1}{2011}\))
= \(\dfrac{2+1}{2}\) \(\times\) \(\dfrac{3+1}{3}\) \(\times\) \(\dfrac{4+1}{4}\)\(\times\) ... \(\times\) \(\dfrac{2010+1}{2010}\)\(\times\) \(\dfrac{2011+1}{2011}\)
= \(\dfrac{3}{2}\)\(\times\)\(\dfrac{4}{3}\)\(\times\)\(\dfrac{5}{4}\)\(\times\)...\(\times\)\(\dfrac{2011}{2010}\)\(\times\)\(\dfrac{2012}{2011}\)
= \(\dfrac{2012}{2}\)
= 1006
Đăt S=1/15+1/35+1/63+1/99+...+1/2915+1/3135
=1/3.5+1/5.7+1/7.9+1/9.11+...+1/53.55+1/55.57
=1/2(2/3.5+2/5.7+2/7.9+...+2/53.55+2/55.57)
=1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/53-1/55+1/55-1/57)
=1/2(1/3-1/57)
=1/2(19/57-1/57)
=1/2.18/57
=3/19
Vậy 1/15+1/35+1/63+1/99+...+1/2915+1/3135=3/19
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Đặt \(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{2915}+\frac{1}{3135}\)
\(\Leftrightarrow A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+....+\frac{1}{53\cdot55}+\frac{1}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{53\cdot55}+\frac{2}{55\cdot57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-....+\frac{1}{53}-\frac{1}{55}+\frac{1}{55}-\frac{1}{57}\)
\(\Leftrightarrow2A=\frac{1}{3}-\frac{1}{57}=\frac{6}{19}\)
\(\Leftrightarrow A=\frac{6}{19}:2=\frac{3}{19}\)
\(\dfrac{8}{30}\) = \(\dfrac{4}{15}\) = \(\dfrac{28}{105}\) < \(\dfrac{33}{105}\)= \(\dfrac{11}{35}\)
vậy 8/30 < 11/35