Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
\(\frac{3}{5}-\frac{-7}{10}-\frac{13}{-20}=\frac{3}{5}+\frac{7}{10}+\frac{13}{20}=\frac{12}{20}+\frac{14}{20}+\frac{13}{20}=\frac{39}{20}\)
\(\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}-\frac{-1}{6}=\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{6+(-4)+3+2}{12}=\frac{7}{12}\)
\(\frac{9}{4}.\frac{8}{27}.\frac{5}{7}=\frac{9.8.5}{4.27.7}=\frac{1.2.5}{1.3.7}=\frac{10}{21}\)
\(\frac{2}{5}.(\frac{2}{3}-\frac{1}{4})+\frac{1}{2}=\frac{2}{5}.(\frac{8}{12}-\frac{3}{12})+\frac{1}{2}=\frac{2}{5}.\frac{5}{12}+\frac{1}{2}=\frac{1}{6}+\frac{1}{2}=\frac{1}{6}+\frac{3}{6}=\frac{4}{6}=\frac{2}{3}\)
\((\frac{1}{3}-\frac{1}{6}):(\frac{1}{3}+\frac{1}{6})=(\frac{2}{6}-\frac{1}{6}):(\frac{2}{6}+\frac{1}{6})=\frac{1}{6}:\frac{3}{6}=\frac{1}{6}.\frac{6}{3}=\frac{1.6}{6.3}=\frac{1.1}{1.3}=\frac{1}{3}\)
Hok tốt
a,15+ {[20/22-5]}
15+{[20/4+5]}
15+{[5+5]}
15+{10+0}
15+10=25
b,15+{[32-4]+2.2}
15+{[9-4]+2.2}
15+{5+2.2}
15+{5+4}
15+9=24
=24
HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
a) - 2 3 .125.32. ( − 76 ) > 0 12.74 . - 3 4 . ( − 395 ) < 0 ⇒ ( − 2 ) 3 .125.32. ( − 76 ) > 12.74. ( − 3 ) 4 . ( − 395 )
b) ( − 1 ) . ( − 2 ) . ( − 3 ) ... ( − 20 ) > 0 ( − 3 ) . ( − 4 ) . ( − 5 ) ... ( − 23 ) < 0 ⇒ ( − 1 ) . ( − 2 ) . ( − 3 ) ... ( − 20 ) > ( − 3 ) . ( − 4 ) . ( − 5 ) ... ( − 23 )
Ta có:
B=1+1/2*(1+2)+1/3*(1+2+3)+..+1/20*(1+2+3+...+20)
B=1+3/2+6/3+10/4+...+210/20
=2/2+3/2+4/2+5/2+...+21/2=115
115 nhé bạn