Chứng minh
a, A = \(\dfrac{1}{5}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{7}\) + ... + \(\dfrac{1}{17}\) < 2
b, B = \(\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{13}\) + ... + \(\dfrac{1}{30}\) > \(\dfrac{2}{3}\)
Những câu hỏi liên quan
chứng minh rằng
A= \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
B=\(\dfrac{5}{11}+\dfrac{5}{12}+\dfrac{5}{13}+\dfrac{5}{14},1< B< 2\)
Bài 1: Thực hiện phép tính:
1) \(\dfrac{-17}{30}-\dfrac{11}{-15}+\dfrac{-7}{12}\)
2) \(\dfrac{-5}{9}+\dfrac{5}{9}:\left(1\dfrac{2}{3}-2\dfrac{1}{12}\right)\)
3) \(\dfrac{-7}{25}.\dfrac{11}{13}+\dfrac{-7}{25}.\dfrac{2}{13}-\dfrac{18}{25}\)
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
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11 tháng 4 2021 lúc 15:06
text{Bài 4. Chứng tỏ rằng:}a) dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{30^2} 1b) dfrac{1}{10}+dfrac{1}{11}+dfrac{1}{12}+...+dfrac{1}{99}+dfrac{1}{100}1c) dfrac{1}{5}+dfrac{1}{6}+dfrac{1}{7}+...+dfrac{1}{17} 2d) dfrac{1}{1.2}+dfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{29.30} 1
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\(\text{Bài 4. Chứng tỏ rằng:}\)
\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)
\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}< 1\)
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
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c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
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Tính hợp lý:
\(A=\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{11}{36}\) \(B=\left(\dfrac{4}{5}+\dfrac{1}{2}\right):\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\)
\(C=\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
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2 tháng 8 2023 lúc 13:49
bài 1)a)(dfrac{12}{42} - dfrac{26}{91} ).(dfrac{-4}{7} +dfrac{-5}{9})b) dfrac{9}{13} . dfrac{-12}{7} +dfrac{9}{13} .dfrac{29}{17} c) dfrac{4}{7} +dfrac{6}{7} .dfrac{1}{7} +dfrac{1}{7} .dfrac{8}{7} d) (dfrac{5}{7} + dfrac{9}{7} ). dfrac{7}{10} +(dfrac{8}{7} + dfrac{13}{7} ). dfrac{7}{10}
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bài 1)
a)(\(\dfrac{12}{42}\) - \(\dfrac{26}{91}\) ).(\(\dfrac{-4}{7}\) +\(\dfrac{-5}{9}\))
b) \(\dfrac{9}{13}\) . \(\dfrac{-12}{7}\) +\(\dfrac{9}{13}\) .\(\dfrac{29}{17}\)
c) \(\dfrac{4}{7}\) +\(\dfrac{6}{7}\) .\(\dfrac{1}{7}\) +\(\dfrac{1}{7}\) .\(\dfrac{8}{7}\)
d) (\(\dfrac{5}{7}\) + \(\dfrac{9}{7}\) ). \(\dfrac{7}{10}\) +(\(\dfrac{8}{7}\) + \(\dfrac{13}{7}\) ). \(\dfrac{7}{10}\)
a: =(2/7-2/7)(-4/7-5/9)=0
b:
Sửa đề: 9/13*(-12/17)+9/13*29/27
=9/13(-12/17+29/17)
=9/13*17/17=9/13
c: \(=\dfrac{1}{7}\left(4+\dfrac{6}{7}+\dfrac{8}{7}\right)=\dfrac{1}{7}\cdot6=\dfrac{6}{7}\)
d: =7/10(5/7+9/7+8/7+13/7)
=5*7/10=7/2
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Kết quả của phép tính \(\dfrac{1}{5}+\dfrac{1}{6}\)bằng?
A. \(\dfrac{13}{30}\)
B. \(\dfrac{17}{30}\)
C. \(\dfrac{37}{30}\)
D. \(\dfrac{11}{30}\)
Xem thêm câu trả lời
a)dfrac{2}{7}+dfrac{-3}{8}+dfrac{11}{7}+dfrac{1}{7}_{ }+dfrac{5}{-8}
b)dfrac{3}{17}+dfrac{-5}{13}+dfrac{-18}{35}+dfrac{14}{17}+dfrac{17}{-35}_{ }+dfrac{-8}{13}
c)dfrac{-3}{8}+dfrac{12}{25}+dfrac{5}{-8}+dfrac{2}{-5}+dfrac{13}{25}
d)dfrac{-5}{21}+left(dfrac{-16}{21}+1right)
e)dfrac{-3}{17}+left(dfrac{2}{3}+dfrac{3}{17}right)
f)left(dfrac{-1}{6}+dfrac{5}{-12}right)+dfrac{7}{12}
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a)\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{7}_{ }+\dfrac{5}{-8}\)
b)\(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}_{ }+\dfrac{-8}{13}\)
c)\(\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
d)\(\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
e)\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
f)\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
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TÍNH:
1)\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
2)\(\dfrac{_{-5}}{11}.\dfrac{13}{17}-\dfrac{5}{11}.\dfrac{4}{17}\)
1,Ta có:\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{57}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\) =\(\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+...+\dfrac{1}{2}\right)\)
= \(\dfrac{9}{10}-\left\{\dfrac{1}{\left(9.10\right)}+\dfrac{1}{\left(9.8\right)}+...+\dfrac{1}{\left(2.1\right)}\right\}\)
= \(\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{1}-\dfrac{1}{2}\right).\left(\dfrac{1}{90}=\dfrac{1}{9.10}=\dfrac{1}{9}-\dfrac{1}{10}\right)\)=\(\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)
=\(\dfrac{9}{10}-\dfrac{9}{10}\)
= 0
Ý 2 dễ rồi bạn tự tính
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1, \(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{6}+\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9.10}+\dfrac{1}{8.9}+...+\dfrac{1}{1.2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{7}-\dfrac{1}{8}+...+1-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{-1}{10}+1\right)=\dfrac{9}{10}-\dfrac{9}{10}=0\)
2, \(\dfrac{-5}{11}\cdot\dfrac{13}{17}-\dfrac{5}{11}.\dfrac{4}{17}\)
\(=\dfrac{-5}{11}\cdot\dfrac{13}{17}+\dfrac{-5}{11}.\dfrac{4}{17}\)
\(=\dfrac{-5}{11}\left(\dfrac{13}{17}+\dfrac{4}{17}\right)=\dfrac{-5}{11}.1=\dfrac{-5}{11}\)
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Tính: a) dfrac{13}{14}-dfrac{-7}{8}+dfrac{-3}{2}b) dfrac{5}{17}+dfrac{-15}{34}.dfrac{2}{5}c) dfrac{1}{5}:dfrac{1}{10}-dfrac{1}{3}.(dfrac{6}{5}-dfrac{2}{4})d) dfrac{-3}{4}:(dfrac{12}{-5}-dfrac{-7}{10})*Lưu ý: Không viết luôn kết quả, giải chi tiết.
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Tính:
a) \(\dfrac{13}{14}\)-\(\dfrac{-7}{8}\)+\(\dfrac{-3}{2}\)
b) \(\dfrac{5}{17}\)+\(\dfrac{-15}{34}\).\(\dfrac{2}{5}\)
c) \(\dfrac{1}{5}\):\(\dfrac{1}{10}\)-\(\dfrac{1}{3}\).(\(\dfrac{6}{5}\)-\(\dfrac{2}{4}\))
d) \(\dfrac{-3}{4}\):(\(\dfrac{12}{-5}\)-\(\dfrac{-7}{10}\))
*Lưu ý: Không viết luôn kết quả, giải chi tiết.
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
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