a) Tính nhanh :
1\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.....1\frac{1}{99}\)
b) CMR : \(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}.....\frac{60}{2}=1.3.5.....59\)
c) Viết PS \(\frac{1}{16}\)thành tổng nghịch đảo của 5 số nguyên \(\ne\)nhau
Những câu hỏi liên quan
Bài 1 : Tính
Cho A =\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+......+\frac{1}{60}>\frac{7}{12}\)
B = \(\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{5^2}+......+\frac{ }{50^{21}}\)
CMR B >\(\frac{1}{4}\)và B < \(\frac{4}{9}\)
C = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.......\frac{79}{80}< \frac{1}{9}\)
CMR:
a, \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b, \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+.....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
CMR
a)\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
CMR:
a) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}<\frac{1}{3}\)
b) \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
a)
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-...-\frac{1}{64}=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-...-\frac{1}{2^6}=A\)
2A = 1 - \(\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^5}\)
2A + A = 1 - \(\frac{1}{2}+\frac{1}{2^2}-...-\frac{1}{2^5}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^6}\)
3A = \(1-\frac{1}{2^6}=\frac{2^6-1}{2^6}\)(đpcm)
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CMR:
a)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}<\frac{1}{3}\)
b)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}<\frac{3}{16}\)
25 tháng 9 2017 lúc 19:28
CMR: \(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}.....\frac{60}{2}=1.3.5.....59\)
\(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)
\(=\)\(\left[\left(31.32.33....60\right)\right]\)\(.\)\(\left(\frac{1.2.3....30}{2^{30}}\right)\)\(.\)\(\left(1.2.3....30\right)\)
\(=\)\(\left[\frac{\left(1.3.5....59\right).\left(2.4.6....60\right)}{2.4.6....60}\right]\)\(=\)\(1.3.5....59\)
Vậy \(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)\(=\)\(1.3.5....59\)
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ta có:Đặt A= \(1.3.5.....59=\frac{1.2.3.4.....59.60}{2.4.6.....60}\)
=\(\frac{1.2.3.....59.60}{2^{30}.\left(1.2.3.....30\right)}=\frac{31.32.....59.60}{2^{30}}\)
= \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
vì \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\) = \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
\(\Rightarrow\)A= \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)
( Điều phải chứng minh)
toán nâng cao lớp 6 đấy bạn nha
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Chứng minh rằng: a)\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Nhanh lên nhé! Mk đang cần gấp.
\(3B=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(B=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4B=3B+B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
+ Đặt \(M=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3M=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow4M=3M+M=3-\frac{1}{3^{99}}\)
\(\Rightarrow M=\frac{3}{4}-\frac{1}{3^{99}\cdot4}\)
\(\Rightarrow4B=M-\frac{100}{3^{100}}=\frac{3}{4}-\frac{1}{3^{99}\cdot4}-\frac{100}{3^{100}}\)
\(\Rightarrow B=\frac{3}{16}-\frac{1}{3^{99}\cdot16}-\frac{100}{3^{100}\cdot4}\) \(\Rightarrow B< \frac{3}{16}\)
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a) \(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
\(\Rightarrow3A=2A+A=1-\frac{1}{2^6}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{2^6\cdot3}< \frac{1}{3}\) ( đpcm )
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Bài 1:So sánh(bằng 2 cách):
\(A=\frac{10^{11}-1}{10^{12}-1}\)\(B=\frac{10^{10}+1}{10^{11}+1}\)
Bài 2:Chứng minh:
\(\frac{3}{5}< \frac{1}{31}+\frac{1}{32}+\frac{1}{33}+..........+\frac{1}{59}+\frac{1}{60}< \frac{4}{5}\)
Câu hỏi của Quỳnh Anh - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo câu 1 2 cách 2 bạn hướng dẫn nhé!
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3.Tính hợp lí:a,frac{17}{5}.frac{-31}{125}.frac{1}{2}.frac{10}{17}.frac{-1}{2^3}b,left(frac{11}{4}.frac{5}{9}-frac{4}{9}.frac{11}{4}right).frac{8}{33}c,left(frac{17}{28}+frac{18}{29}-frac{19}{30}-frac{20}{31}right).left(frac{-5}{12}+frac{1}{4}+frac{1}{6}right)4.Tìm tích:a,left(frac{1}{2}+1right)left(frac{1}{3}+1right).....left(frac{1}{99}+1right)b,left(frac{1}{2}-1right)left(frac{1}{3}-1right).....left(frac{1}{100}-1right)c,frac{3}{2^2}.frac{8}{3^2}.frac{15}{4^2}....frac{899}{30^2}AI CÒN THỨC TH...
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3.Tính hợp lí:
a,\(\frac{17}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}\)
b,\(\left(\frac{11}{4}.\frac{5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\)
c,\(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
4.Tìm tích:
a,\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right).....\left(\frac{1}{99}+1\right)\)
b,\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).....\left(\frac{1}{100}-1\right)\)
c,\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}....\frac{899}{30^2}\)
AI CÒN THỨC THÌ GIÚP MIK VS,MIK ĐANG CẦN GẤP
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} đây là biểu thức gì\)