Chứng tỏ rằng: \(\frac{373}{420}=\frac{373373373}{420420420}\)
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Cho M=\(\frac{1}{10}+\frac{1}{15}+\frac{1}{20}+\frac{1}{28}+...+\frac{1}{105}+\frac{1}{420}\)chứng minh rằng \(\frac{1}{3}< M< \frac{1}{2}\)
Cho S =\(\frac{1}{50}\)+\(\frac{1}{51 }\)+\(\frac{1}{52}\)+...+\(\frac{1}{98}\)+\(\frac{1}{99}\)
Chứng tỏ rằng S >\(\frac{1}{2}\)
DDODOGDOGE
Giải:
\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\)
\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\)
\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\)
\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\)
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Ta có:S=1/50+1/51+1/52+...+1/99
S>1/50+1/50+1/50+....+1/50(50 số hạng)
S>1/50x50
S>1>1/2
=>S>1/2
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bài 1,Cho \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\).Hãy chứng tỏ rằng B>1.
bài 2,Cho \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\).Hãy chứng tỏ rằng S<1.
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)
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Cho Afrac{1}{101}+frac{1}{102}+frac{1}{103}+ . . . . . . . .+frac{1}{199}+frac{1}{200}a)chứng tỏ rằng Afrac{3}{4}(tách 2 dòng) chứng tỏ rằng A frac{7}{12}(tách 2 dòng) chứng tỏ rằng A frac{5}{8}(tách 4 dòng)
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Cho A=\(\frac{1}{101}\)+\(\frac{1}{102}\)+\(\frac{1}{103}\)+ . . . . . . . .+\(\frac{1}{199}\)+\(\frac{1}{200}\)
a)chứng tỏ rằng A<\(\frac{3}{4}\)(tách 2 dòng)
chứng tỏ rằng A >\(\frac{7}{12}\)(tách 2 dòng)
chứng tỏ rằng A >\(\frac{5}{8}\)(tách 4 dòng)
a) Chứng tỏ rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) Đặt A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013+2014}\); Đặt B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
Chứng tỏ rằng \(\frac{A}{B}\)là số nguyên
a) Chứng tỏ rằng
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) Đặt A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\); B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
Chứng tỏ rằng \(\frac{A}{B}\)là số nguyên
Chứng tỏ rằng:
$\frac{k}{n.(n +k)}$ = $\frac{1}{n}$ - $\frac{1}{n + k}$
Chứng tỏ rằng:
$\frac{k}{n.(n + k)}$ = $\frac{1}{n}$ - $\frac{1}{n + k}
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Chứng tỏ rằng :1.5.7...197.199=\(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}...\frac{200}{2}\)
#)Giải :
Ta có : \(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}.....\frac{200}{2}=\frac{101.102.103.....200}{2^{100}}=\frac{\left(101.102.103.....200\right)\left(1.2.3.....100\right)}{2^{100}\left(1.2.3.....100\right)}\)
\(=\frac{1.2.3.....200}{\left(2.1\right)\left(2.2\right)\left(2.3\right)...\left(2.100\right)}=\frac{\left(1.3.5.....99\right)\left(2.4.6.....100\right)}{2.4.6.....200}=1.3.5.....99\left(đpcm\right)\)
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Ta có : 1.3.5.7.....199 = \(\frac{\left(1.3.5.7.....199\right).\left(2.4.6.8.....200\right)}{2.4.6.8.....200}=\frac{1.2.3.4.5.....199.200}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(100.2\right)}=\frac{1.2.3.4.5.....199.200}{2^{100}.1.2.3.....100}=\frac{101.102.103.....200}{2^{100}}\)\(=\frac{101}{2}.\frac{102}{2}\frac{103}{2}.....\frac{200}{2}\)\( \left(ĐPCM\right)\)
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6 tháng 2 2020 lúc 11:48
chứng tỏ rằng : A=\(\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+....+\frac{36}{25.27.29}< 3\)
Ta có:
\(A=\frac{36}{1.3.5}+\frac{36}{3.5.7}+\frac{36}{5.7.9}+...+\frac{36}{25.27.29}\)
\(\Rightarrow A=9.\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{1.3}-\frac{1}{27.29}\right)\)
\(\Rightarrow A=9.\left(\frac{1}{3}-\frac{1}{783}\right)\)
\(\Rightarrow A=9.\frac{1}{3}-9.\frac{1}{783}\)
\(\Rightarrow A=3-\frac{1}{87}\)
Vì \(3-\frac{1}{87}< 3.\)
\(\Rightarrow A< 3\left(đpcm\right).\)
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