A=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+.....+\frac{1}{70}\)
CM\(\frac{4}{3}< A< \frac{46}{15}\)
Những câu hỏi liên quan
Cho A=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}.Chungminhrang\frac{4}{3}< A< \frac{5}{2}\)
Bài 1 :Thực hiện phép tính :
a) M (frac{-6}{13}+frac{15}{26}-frac{47}{39}-frac{1}{78}) : (99frac{17}{65}-100frac{5}{52}+frac{1}{130})
b) N frac{(frac{3}{5}-0,435+frac{1}{200}):left(-0,04right)}{30,75+frac{1}{12}+3frac{1}{6}}
c) P (frac{-5}{6}:frac{-10}{11})+frac{frac{1}{4}+frac{5}{8}-frac{7}{13}}{frac{-2}{12}-frac{10}{24}+frac{14}{39}}
Bài 2 : Thực hiện phép tính :V
a) P frac{frac{1}{5}-frac{1}{9}+frac{1}{13}}{frac{9}{5}-1+frac{9}{13}}+frac{frac{10}{7}-frac{10}{11}-frac{10}{17}}{frac{12}{...
Đọc tiếp
Bài 1 :Thực hiện phép tính :
a) M =(\(\frac{-6}{13}+\frac{15}{26}-\frac{47}{39}-\frac{1}{78}\)) : (\(99\frac{17}{65}-100\frac{5}{52}+\frac{1}{130}\))
b) N = \(\frac{(\frac{3}{5}-0,435+\frac{1}{200}):\left(-0,04\right)}{30,75+\frac{1}{12}+3\frac{1}{6}}\)
c) P = (\(\frac{-5}{6}:\frac{-10}{11}\))+\(\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\)
Bài 2 : Thực hiện phép tính :V
a) P =\(\frac{\frac{1}{5}-\frac{1}{9}+\frac{1}{13}}{\frac{9}{5}-1+\frac{9}{13}}+\frac{\frac{10}{7}-\frac{10}{11}-\frac{10}{17}}{\frac{12}{7}-\frac{12}{11}-\frac{12}{17}}\)
b) Q = \(\frac{\frac{1}{14}-\frac{1}{30}-\frac{1}{46}}{\frac{2}{35}-\frac{2}{75}-\frac{2}{115}}:\frac{\frac{3}{8}-\frac{15}{17}+\frac{30}{31}}{\frac{1}{6}-\frac{20}{51}+\frac{40}{93}}\)
có rất nhiều câu dễ ở trong đề sao bạn Ko thử làm đi rồi câu nào khó lại hỏi
A=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{70}\)
Chứng minh rằng:\(\frac{4}{3}< A< 35\)
Tìm x:
a) (2,8x - 32) :\(\frac{2}{3}\)=90
b) 4,5 - 2x . \(1\frac{4}{7}\)=\(\frac{11}{14}\)
c) (x+ \(\frac{1}{4}-\frac{1}{3}\) ): (2+\(\frac{1}{6}-\frac{1}{4}\))=\(\frac{7}{46}\)
d)\(\frac{13}{15}-\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{7}{10}\)
\(\left(2.8x-32\right):\frac{2}{3}=90\)
\(2.8\cdot x-32=90\cdot\frac{2}{3}\)
\(\frac{14}{5}x-32=60\)
\(\frac{14}{5}x=60+32\)
\(\frac{14}{5}x=92\)
\(x=\frac{230}{7}\)
B , c , d tương tự
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A=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)
CMR \(\frac{4}{3}\) bé hơn A bé hơn 2,5.
Làm nhanh giùm mình!!!!!
để chứng minh A > \(\frac{4}{3}\)ta tách tổng A thành 3 nhóm :
A = \(\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{70}\right)\)
A > \(\frac{1}{30}.20+\frac{1}{50}.20+\frac{1}{70}.20=\frac{2}{3}+\frac{2}{5}+\frac{2}{7}=1\frac{37}{105}>1\frac{35}{105}=1\frac{1}{3}=\frac{4}{3}\)
để chứng minh A < 2,5 ta tách tổng A thành 6 nhóm :
A = \(\left(\frac{1}{11}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+...+\frac{1}{70}\right)\)
A < \(\frac{1}{11}.10+\frac{1}{21}.10+\frac{1}{31}.10+\frac{1}{41}.10+\frac{1}{51}.10+\frac{1}{61}.10< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{4}+\frac{1}{5}\right)< 2+0,5=2,5\)
Bạn có hiểu không chi le hay để mình giải thích cho
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Ta tách biểu thức thành 7 nhóm , t CÓ các nhóm sau :
- \(\frac{1}{11}\)+\(\frac{1}{12}\)+\(\frac{1}{13}\)+...+\(\frac{1}{20}\)
- .....
Ta thấy tất cả các phân số trên đều > hơn \(\frac{1}{20}\)
=> \(\frac{1}{11}\)+\(\frac{1}{12}\)+\(\frac{1}{13}\)+....+\(\frac{1}{20}\)> \(\frac{10}{20}\)=\(\frac{1}{2}\) ( VÌ CÓ 10 phân số đều lớn hơn hoặc = \(\frac{1}{20}\))
Tương tự với 7 nhóm còn lại mỗi nhóm gồm 10 phân số ta được các phân số \(\frac{1}{3}\),\(\frac{1}{4}\),\(\frac{1}{5},\frac{1}{6},\frac{1}{7}\)
Ta cộng tổng các p/s \(\frac{1}{3},\frac{1}{4}\frac{1}{5},\frac{1}{6},\frac{1}{7}\)ta được p/s \(\frac{223}{140}>\frac{4}{3}\)
=> ĐIỀU PHẢI CHỨNG MINH
Mk chỉ làm được ở chỗ 4/3 < A thôi
Vậy nhé bạn yêu wys!!!!!!!!!!!!!!
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a)frac{15}{12}+frac{5}{13}-frac{3}{12}-frac{18}{13}b)frac{11}{24}-frac{5}{41}+frac{13}{24}+0,5-frac{36}{41}c)left(-frac{3}{4}+frac{2}{3}right):frac{5}{11}+left(-frac{1}{4}+frac{1}{3}right):frac{5}{11}d)left(-3right)^2cdotleft(frac{3}{4}-0,25right)-left(3frac{1}{2}-1frac{1}{2}right)e)frac{13}{25}+frac{6}{41}-frac{38}{25}+frac{35}{41}-frac{1}{2}
Đọc tiếp
a)\(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}\)
b)\(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)
c)\(\left(-\frac{3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(-\frac{1}{4}+\frac{1}{3}\right):\frac{5}{11}\)
d)\(\left(-3\right)^2\cdot\left(\frac{3}{4}-0,25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)
e)\(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
a) \(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
b) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}=\left(\frac{11}{24}+\frac{13}{24}\right)+\left(-\frac{5}{41}-\frac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
_Học tốt nha_
a, \(\frac{15}{12}\)+ \(\frac{5}{13}\)- \(\frac{3}{12}\)-\(\frac{18}{13}\)
= \(\frac{5}{4}\)+ \(\frac{5}{13}\) - \(\frac{1}{4}\) - \(\frac{18}{13}\)
= \(\left(\frac{5}{4}-\frac{1}{4}\right)\)+ \(\left(\frac{5}{13}-\frac{18}{13}\right)\)
= 1 - 1 = 0
b, \(\frac{11}{24}\)- \(\frac{5}{41}\)+ \(\frac{13}{24}\)+ 0,5 - \(\frac{36}{41}\)
= \(\left(\frac{11}{24}+\frac{13}{24}\right)\)- \(\left(\frac{5}{41}+\frac{36}{41}\right)\)+ 0,5
= 1 - 1 + 0,5 = 0,5
c, \(\left(-\frac{3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(-\frac{1}{4}+\frac{1}{3}\right):\frac{5}{11}\)
=\(\left(-\frac{3}{4}+\frac{2}{3}\right).\frac{11}{5}+\left(-\frac{1}{4}+\frac{1}{3}\right).\frac{5}{11}\)
= \(\frac{11}{5}.\left(-\frac{3}{4}+\frac{2}{3}-\frac{1}{4}+\frac{1}{3}\right)\)
= \(\frac{11}{5}.\left[\left(-\frac{3}{4}-\frac{1}{4}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)\right]\)
= \(\frac{11}{5}.\left[\left(-1\right)+1\right]\)
= 0
d, \(\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)
= \(9.\left(0,75-0,25\right)-2\)
= 9. 0,5 - 2 = 2,5
e, \(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
= \(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
= -1 + 1 - \(\frac{1}{2}\)
= \(-\frac{1}{2}\)
a) 15/12 + 5/13- 3/12 - 18/13 = (15/12 - 3/12) + ( 5/13 - 18/13) = 12/12 + -13/13 = 1 + (-1) = 0
b) 11/24 - 5/41 + 13/24 + 0,5 - 36/41 = (11/24 +13/24) - (5/41+36/41)+0,5 = 1 - 1+0,5 = 0,5
c) ( -3/4 + 2/3) : 5/11 + (-1/4 + 1/3 ) : 5/11 = -3/4 + 2/3 : 5/11 + -1/4 + 1/3 = = [( -3/4 + (-1/4) ] + ( 2/3 + 1/3) : 5/11
= -4/4 + 3/3 : 5/11 = -1 + 1 * 11/5
= 0 * 11/5 = 0
d) (-3) ^2 * (3/4 - 0,25) - ( 3 1/2 - 1 1/2) = 9 * (3/4 - 25/100) - ( 7/2 -3/2) = 9 * ( 3/4 - 1/4) - 4/2
= 9* 1/2 - 2 = 9/ 2 - 2= 5/2
e) 13/25 + 6/41 - 38/25 + 35/41 - 1/2 = ( 13/25 - 38/25) + ( 6/41 + 35/41) - 1/2 = -25/25 + 41/41 - 1/2 = (-1) + 1 - 1/2 = 0 - 1/2 = -1/2
tìm\(x\in z\)biết :
a.\(\frac{13}{15}-\left(\frac{13}{21}+x\right)\times\frac{7}{12}=\frac{7}{10}\)
b.\(\left(x+\frac{1}{4}-\frac{1}{3}\right)\div\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
Thực hiện phép tính:
a) \(\frac{1}{12}+\frac{3}{15}+\frac{11}{12}+\frac{1}{71}-\frac{12}{10}\)
b) \(\frac{2}{3}-4.\left(\frac{1}{2}+\frac{3}{4}\right)\)
c) \(\frac{-4}{13}.\frac{5}{17}+\frac{-12}{13}.\frac{4}{7}+\frac{4}{13}\)
d) \(\frac{10^3+2.5+5^3}{55}\)
a) \(\frac{1}{12}+\frac{3}{15}+\frac{11}{12}+\frac{1}{71}-\frac{12}{10}=\left(\frac{1}{12}+\frac{11}{12}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\frac{1}{71}\)
\(=\frac{12}{12}+0+\frac{1}{71}=1+\frac{1}{71}=1\frac{1}{71}=\frac{72}{71}\)
b) \(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)=\frac{2}{3}-4.\frac{5}{4}=\frac{2}{3}-5=\frac{2}{3}-\frac{15}{3}=-\frac{13}{3}\)
c) \(\frac{-4}{13}.\frac{3}{17}+\frac{-12}{13}.\frac{4}{7}+\frac{4}{13}=\frac{4}{13}.\frac{-3}{17}+\frac{4}{13}.\frac{-12}{17}+\frac{4}{13}.1\)
\(=\frac{4}{13}\left(\frac{-3}{17}+\frac{-12}{17}+1\right)=\frac{4}{13}\left(\frac{-15}{17}+\frac{17}{17}\right)=\frac{4}{13}.\frac{2}{17}=\frac{8}{221}\)
d) \(\frac{10^3+2.5+5^3}{55}=\frac{1000+10+125}{55}=\frac{1135}{55}=\frac{227}{11}\)
Bài 1:
a) A = 1 +\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\) . Chứng minh rằng A \(⋮\) 100.
b) A = \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\). Chứng minh rằng A > \(\frac{4}{3}\)
b
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+..+\frac{1}{70}\)
Ta thấy:
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)( có 10 phân số \(\frac{1}{20}\)) = \(\frac{1}{20}\).10 = \(\frac{1}{2}\)
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\)(có 10 phân số \(\frac{1}{30}\)) = \(\frac{1}{30}\).10 = \(\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( có 10 phân số \(\frac{1}{40}\)) = \(\frac{1}{40}\).10 = \(\frac{1}{4}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\)( có 10 phân số \(\frac{1}{50}\)) =\(\frac{1}{50}.10=\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)( có 10 phân số \(\frac{1}{60}\)) =\(\frac{1}{60}.10=\frac{1}{6}\)
\(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}>\frac{1}{70}+\frac{1}{70}+...+\frac{1}{70}\)( có 10 phân số \(\frac{1}{70}\)) \(=\frac{1}{70}.10=\frac{1}{7}\)
=> A> \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\frac{223}{140}=\frac{699}{420}>\frac{560}{420}=\frac{4}{3}\)
=> A > \(\frac{4}{3}\)
có bài toán nào khó thì ib mk nha
a)
\(A=1+\frac{1}{2}+\frac{1}{3}+..+\frac{1}{99}\)
\(A=\left(1+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{98}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)+\frac{1}{50}\)
\(A=\frac{100}{99}+\frac{100}{98.2}+...+\frac{100}{49.51}+\frac{1}{50}\)
\(A=100\left(\frac{1}{99}+\frac{1}{98.2}+...+\frac{1}{49.51}\right)+\frac{1}{50}\)
Ta Thấy \(100\left(\frac{1}{99}+\frac{1}{98.2}+...+\frac{1}{49.51}\right)⋮100\)mà \(\frac{1}{50}\)\(⋮̸\)100
=> A \(⋮̸\) 100
Nếu đề bài là \(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{51}+...+\frac{1}{99}\)thì bạn áp dụng cách tính bên trên của mk là ra hem