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a, Ta có :
\(\dfrac{1}{6}< \dfrac{1}{5}\)
\(\dfrac{1}{7}< \dfrac{1}{5}\)
.................
\(\dfrac{1}{9}< \dfrac{1}{5}\)
\(\dfrac{1}{10}=\dfrac{1}{10}\)
\(\dfrac{1}{11}< \dfrac{1}{10}\)
..................
\(\dfrac{1}{17}< \dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+......+\dfrac{1}{17}< \dfrac{1}{5}+\dfrac{1}{5}+....+\dfrac{1}{5}\)
\(\Leftrightarrow A< \dfrac{1}{5}.5+\dfrac{1}{10}.8\)
\(\Leftrightarrow A< 1+\dfrac{4}{5}=\dfrac{9}{5}< 2\)
\(\Leftrightarrow A< 2\left(đpcm\right)\)
b/ Ta có :
\(\dfrac{1}{11}>\dfrac{1}{30}\)
\(\dfrac{1}{12}>\dfrac{1}{30}\)
...............
\(\dfrac{1}{29}>\dfrac{1}{30}\)
\(\dfrac{1}{30}=\dfrac{1}{30}\)
\(\Leftrightarrow\dfrac{1}{11}+\dfrac{1}{12}+........+\dfrac{1}{30}>\dfrac{1}{30}+\dfrac{1}{30}+.......+\dfrac{1}{30}\)
\(\Leftrightarrow B>\dfrac{1}{30}.20=\dfrac{2}{3}\)
\(\Leftrightarrow B>\dfrac{2}{3}\left(đpcm\right)\)
Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-1-\dfrac{1}{2}-...-\dfrac{1}{10}\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
Vậy \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
\(=\left(\dfrac{1}{10}+\dfrac{-1}{10}\right)+\left(-\dfrac{1}{11}+\dfrac{1}{11}\right)+\left(-\dfrac{1}{12}+\dfrac{1}{12}\right)+\left(-\dfrac{1}{13}+\dfrac{1}{13}\right)+\left(-\dfrac{1}{14}+\dfrac{1}{14}\right)+\left(-\dfrac{1}{15}+\dfrac{1}{15}\right)+\dfrac{1}{16}\\ =\dfrac{1}{16}\)
Tính nhanh :
\(\dfrac{1}{10}+\dfrac{-1}{11}+\dfrac{1}{12}+\dfrac{-1}{13}+\dfrac{1}{14}+\dfrac{-1}{15}+\dfrac{1}{16}+\dfrac{-1}{10}+\dfrac{1}{11}+\dfrac{-1}{12}+\dfrac{1}{13}+\dfrac{-1}{14}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{10}+\dfrac{-1}{10}\right)+\left(\dfrac{-1}{11}+\dfrac{1}{11}\right)+\left(\dfrac{1}{12}+\dfrac{-1}{12}\right)+\left(\dfrac{-1}{13}+\dfrac{1}{13}\right)+\left(\dfrac{1}{14}+\dfrac{-1}{14}\right)\)
\(+\left(\dfrac{-1}{15}+\dfrac{1}{15}\right)+\dfrac{1}{16}\)
\(=0+0+...+0+\dfrac{1}{16}\)
\(=\dfrac{1}{16}\)
a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)
b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)
c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)
d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)
lưu ý mk ko chép đầu bài
mình cần gấp lắm đến chiều mai là phải nộp rùi
giúp mình nha thanks cá bạn trước
ko có tâm trạng mà cười nữa![]()
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2)
S = \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{43.46}\)
S = 3 . (\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{43.46}\))
S = 1 . (\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{43.46}\))
S = 1 . (\(1-\dfrac{1}{4}+...+\dfrac{1}{43}-\dfrac{1}{46}\))
S = 1 . (\(1-\dfrac{1}{46}\))
S = 1 . \(\dfrac{45}{46}\)
S = \(\dfrac{45}{46}\)
=> \(\dfrac{45}{46}\) < 1
Tối được không bạn?
Xin lỗi nha trễ tí
Trang 2 nè nhìn lại cho rõ nha
giúp mk với mọi người lm ơn đó
ok, từ h đến chiều mai lúc nào cũng đc, cảm ơn bạn nhé mk hứa sẽ tích đúng cho
Uhm bạn
thks
Chưa giải mà đã cảm ơn rồi :v Lỡ giải sai thì sao
thì thôi
mình viết nhầm đề đấy, thật ra bài này mình lm đc rồi mà mk nhìn nhầm viết luôn vào, nhưng cúng cảm ơn bạn nhé, mk tick đúng cho bạn r