Cho B = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}+\frac{1}{9^2}\) .Chứng minh: \(\frac{2}{5}< B< \frac{8}{9}\)
các bạn ơi! giúp mình với
Những câu hỏi liên quan
a)\(\frac{7}{x}<\frac{x}{4}<\frac{10}{x}\)
b) Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\). Chứng tỏ: \(\frac{8}{9}>A>\frac{2}{5}\)
Giải:
a) \(\dfrac{7}{x}< \dfrac{x}{4}< \dfrac{10}{x}\)
\(\Rightarrow7< \dfrac{x^2}{4}< 10\)
\(\Rightarrow\dfrac{28}{4}< \dfrac{x^2}{4}< \dfrac{40}{4}\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6\)
b) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{8}{9}\left(1\right)\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
\(...\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\left(2\right)\)
Từ (1) và (2), ta có:
\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\left(đpcm\right)\)
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Bài 1: a) Cho bin n:b1. Chứng minh rằng: frac{1}{b}-frac{1}{b+1} frac{1}{b^2}-frac{1}{b-1}-frac{1}{b}(1)b) Áp dụng công thức (1) chứng minh frac{2}{5} frac{1}{2^2}+frac{1}{3^2}+frac{1}{4^2}+...+frac{1}{8^2}+frac{1}{9^2} frac{8}{9}Bài 2. Chứng tỏfrac{1}{1.5}+frac{1}{5.9}+frac{1}{9.13}+...+frac{1}{21.25} frac{1}{4}
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Bài 1:
a) Cho \(b\in n\):\(b>1\). Chứng minh rằng: \(\frac{1}{b}-\frac{1}{b+1}< \frac{1}{b^2}-\frac{1}{b-1}-\frac{1}{b}\)(1)
b) Áp dụng công thức (1) chứng minh \(\frac{2}{5}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}+\frac{1}{9^2}< \frac{8}{9}\)
Bài 2. Chứng tỏ
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}< \frac{1}{4}\)
Các bạn ơi ,giúp mình vớiBài 1:Rút gọna)frac{2^{19}times27^3+15times4^9times9^4}{6^9times2^{10}+12^{10}}b)frac{left(frac{-1}{2}right)^3-left(frac{3}{4}right)^3timesleft(-2right)^2}{2timesleft(-1right)^5+left(frac{3}{4}right)^2-frac{3}{8}}c)frac{45times9^4-2times6^4}{2^{19}times3^8+6^8times20}Bài 2:Tìm xa)5^x+5^{x+2}650b)3^{x-1}+5times3162
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Các bạn ơi ,giúp mình với
Bài 1:Rút gọn
a)\(\frac{2^{19}\times27^3+15\times4^9\times9^4}{6^9\times2^{10}+12^{10}}\)
b)\(\frac{\left(\frac{-1}{2}\right)^3-\left(\frac{3}{4}\right)^3\times\left(-2\right)^2}{2\times\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}}\)
c)\(\frac{45\times9^4-2\times6^4}{2^{19}\times3^8+6^8\times20}\)
Bài 2:Tìm x
a)\(5^x+5^{x+2}=650\)
b)\(3^{x-1}+5\times3=162\)
Giúp mình với các bạn, mình đang cần gấp xin các bạn giúp mình nhé
\(\frac{1}{2}.\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right);\left(\frac{-5}{28}+1,75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right);\frac{1}{3}.\frac{5}{7}-\frac{7}{27}.\frac{36}{14};70,5-528:\frac{15}{2}\)
A) \(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\)
\(=\frac{1}{4}\)
B) \(\left(\frac{-5}{28}+1.75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(=\left(\frac{-5}{28}+\frac{7}{4}+\frac{8}{35}\right):\frac{-69}{20}\)
\(=\frac{14}{5}:\frac{-69}{20}\)
\(=\frac{-56}{69}\)
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cho \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{9^2}\)
chứng minh \(\frac{2}{5}\)<A<\(\frac{8}{9}\)
bài 1:
\(a,\frac{1}{3}-\left(\frac{1}{2}+\frac{1}{8}\right)\)
\(b,\frac{1}{2}-\frac{1}{4}+\frac{1}{13}+\frac{1}{8}\)
\(c,13\frac{2}{7}:\left(\frac{-8}{9}\right)+2\frac{5}{7}:\left(\frac{-8}{9}\right)\)
Các bạn làm giúp mk nha
Làm đc phần nào thì làm
Ko cần làm hết 3 phần cũng đc
a) \(\frac{1}{3}-\left(\frac{1}{2}+\frac{1}{8}\right)\)
= \(\frac{1}{3}-\left(\frac{4}{8}+\frac{1}{8}\right)\)
= \(\frac{1}{3}-\frac{5}{8}\)
= \(\frac{8}{24}-\frac{15}{24}\)
= \(\frac{-7}{24}\)
b) \(\frac{1}{2}-\frac{1}{4}+\frac{1}{13}+\frac{1}{8}\)
= \(\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\right)\)+ \(\frac{1}{13}\)
= \(\left(\frac{4}{8}-\frac{2}{8}+\frac{1}{8}\right)+\frac{1}{13}\)
= \(\frac{1}{8}+\frac{1}{13}\)
= \(\frac{13}{104}+\frac{8}{104}\)
= \(\frac{23}{104}\)
c) \(13\frac{2}{7}:\left(\frac{-8}{9}\right)+2\frac{5}{7}:\left(\frac{-8}{9}\right)\)
= \(\left(13\frac{2}{7}+2\frac{5}{7}\right):\left(\frac{-8}{9}\right)\)
= \(16:\left(\frac{-8}{9}\right)\)
= -18
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1.Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^3.4^2}+...+\frac{19}{9^2.10^2}< 1\)
2.Chứng minh rằng: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Làm nhanh giúp mình nhé mọi người !!!
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
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Có phải ở sách NCPT ko bn
Bài 2: Đặt \(B=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(3B=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(3B-B=\left(1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\right)-\left(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\right)\)
\(2B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(6B=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(6B-2B=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\right)\)
\(4B=3-\frac{100}{3^{99}}-\frac{1}{3^{99}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{300}{3^{100}}-\frac{3}{3^{100}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{303}{3^{100}}+\frac{100}{3^{100}}\)
\(4B=3-\frac{203}{3^{100}}< 3\)
\(B< \frac{3}{4}\left(đpcm\right)\)
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Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)Chứng minh \(\frac{2}{5}< A< \frac{8}{9}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow A< 1-\frac{1}{9}=\frac{8}{9}\)(1)
Lại có: \(A>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\Rightarrow A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(A>\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)(2)
Từ (1) và (2), suy ra: \(\frac{2}{5}< A< \frac{8}{9}\)
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Mấy bạn giúp mình vớiBài 1:Chứng minh rằng:a)frac{1}{1}-frac{1}{2}+frac{1}{3}-frac{1}{4}+...+frac{1}{9}-frac{1}{10}frac{1}{6}+frac{1}{7}+frac{1}{8}+frac{1}{9}+frac{1}{10}b)frac{1}{1times2}+frac{1}{3times4}+frac{1}{5times6}+...+frac{1}{400}frac{1}{201}+frac{1}{202}+...+frac{1}{400}
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Mấy bạn giúp mình với
Bài 1:Chứng minh rằng:
a)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
b)\(\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+...+\frac{1}{400}=\frac{1}{201}+\frac{1}{202}+...+\frac{1}{400}\)