Chứng minh:B=\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113}< \dfrac{1}{2}\)
Giúp mik vs cb ! Tối nay mik đi học rồi !
Những câu hỏi liên quan
Bài 1: Chứng tỏ rằng :
dfrac{11}{15} dfrac{1}{21}+dfrac{1}{22}+......+dfrac{1}{60} dfrac{3}{2}
Bài 2: Chứng tỏ rằng:
dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+......+dfrac{1}{n^2} 1
dfrac{1}{4}+dfrac{1}{16}+dfrac{1}{36}+dfrac{1}{100}+dfrac{1}{144}+dfrac{1}{196} dfrac{1}{2}
dfrac{1}{5}+dfrac{1}{13}+dfrac{1}{25}+dfrac{1}{41}+dfrac{1}{61}+dfrac{1}{85}+dfrac{1}{113} dfrac{1}{2}
Đọc tiếp
Bài 1: Chứng tỏ rằng :
\(\dfrac{11}{15}< \dfrac{1}{21}+\dfrac{1}{22}+......+\dfrac{1}{60}< \dfrac{3}{2}\)
Bài 2: Chứng tỏ rằng:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+......+\dfrac{1}{n^2}< 1\)
\(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}< \dfrac{1}{2}\)
\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113}< \dfrac{1}{2}\)
bài 2
| a;đặt biểu thức là S | |
| S < 1/1.2 + 1/2.3 + .......1/(n-1)n | |
| = 1- 1/2 +1 /2 -1/3+........ + 1/n-1 - 1/n | |
|
= 1 -1/n <1 |
|
| vậy S < 1 | |
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So sánh: A=\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{37}+\dfrac{1}{61}+\dfrac{1}{77}+\dfrac{1}{113}\)và B=\(\dfrac{1}{2}\)
So sánh: A=\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{37}+\dfrac{1}{61}+\dfrac{1}{77}+\dfrac{1}{113}\)và B=\(\dfrac{1}{2}\)
Xem thêm câu trả lời
Tìm \(x\):
a)\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\) b)\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
giúp mik vs ạ mik cần gấp
a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh
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a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x-\dfrac{23}{14}=1\)
\(x=1+\dfrac{23}{14}\)
\(x=\dfrac{37}{14}\)
b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{28}{15}\)
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Chứng tỏ rằng:
a) Sdfrac{1}{5}+dfrac{1}{13}+dfrac{1}{14}+dfrac{1}{15}+dfrac{1}{61}+dfrac{1}{62}+dfrac{1}{63} dfrac{1}{2}
b) Sdfrac{1}{41}+dfrac{1}{42}+...+dfrac{1}{80}dfrac{7}{12}
c) Sdfrac{1}{2}+dfrac{1}{2^2}+dfrac{1}{2^3}+...+dfrac{1}{2^{20}} 1
d) dfrac{49}{100} Sdfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{99^2} 1
Các bạn giải ra từng bước dùm mik nha
Thanks m.n
Đọc tiếp
Chứng tỏ rằng:
a) \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{2}\)
b) \(S=\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{80}>\dfrac{7}{12}\)
c) \(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{20}}< 1\)
d) \(\dfrac{49}{100}< S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}< 1\)
Các bạn giải ra từng bước dùm mik nha
Thanks m.n
\(\left(\dfrac{-2}{5}\right)^2\) . \(|\dfrac{1}{3}-\dfrac{3}{5}|\) -\(\dfrac{2}{5}\) \(\sqrt{\dfrac{1}{25}}\) + 1\(\dfrac{1}{3}\)
Tính giúp mik nha, mik đang cần gấp
\(\left(-\dfrac{2}{5}\right)^2\cdot\left|\dfrac{1}{3}-\dfrac{3}{5}\right|-\dfrac{2}{5}\cdot\sqrt{\dfrac{1}{25}}+\dfrac{4}{3}\)
\(=\dfrac{4}{25}\cdot\dfrac{4}{15}-\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{4}{3}\)
\(=\dfrac{16}{375}-\dfrac{2}{25}+\dfrac{4}{3}\)
\(=\dfrac{16}{375}-\dfrac{30}{375}+\dfrac{500}{375}\)
\(=\dfrac{486}{375}=\dfrac{162}{125}\)
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7 tháng 7 2021 lúc 16:51
Bài 3:
c) Chứng tỏ rằng \(\dfrac{1}{15}+\dfrac{1}{16}+\dfrac{1}{17}+...+\dfrac{1}{43}+\dfrac{1}{44}>\dfrac{5}{6}\)
Giúp mik vs! Thanks nhiều nha!
Tìm \(x\) là số tự nhiên biết:
a)\(\dfrac{2}{3}+\dfrac{3}{4}< x< 1\dfrac{1}{3}+\dfrac{4}{5}\) b)\(\dfrac{5}{6}-\dfrac{1}{4}< x< 2\dfrac{1}{3}-\dfrac{2}{5}\)
mn giúp mik vs mik cần gấp
`a, 2/3 +3/4 = (8+9)/12=17/12.`
`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.
`=> x=2.`
`b, 5/6-1/4=(20-6)/24=7/12`.
`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.
`=> x=1`.
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a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)
-> 1 1/3 + 4/5 = 4/3 + 4/5 = 20+12/15 = 32/15
vậy x có thể = 14/14 = 1 (x thuộc N)
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Chứng minh rằng:
\(\dfrac{1}{5}-\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\)
Sai đề. Sửa đề :v
Cmr: \(\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\)
Giải:
Đặt \(A=\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}\)
Ta có:
\(A=\dfrac{1}{5}+\left(\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}\right)+\left(\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}\right)\)
\(A< \dfrac{1}{5}\left(\dfrac{1}{14.3}\right)+\left(\dfrac{1}{61.3}\right)\)
\(A< \dfrac{1}{5}+\dfrac{3}{14}+\dfrac{3}{61}\)
\(A< \dfrac{1}{5}+\dfrac{3}{12}+\dfrac{1}{20}\)
\(A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)
\(\Rightarrow A< \dfrac{1}{2}\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{14}+\dfrac{1}{28}+\dfrac{1}{44}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{97}< \dfrac{1}{2}\) \((đpcm)\)
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