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A = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2
1/2^2 < 1/1*2
1/3^2 < 1/2*3
1/4^2 < 1/3*4
...
1/100^2 < 1/99*100
=> A < 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/99*100
=> A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
=> A < 1 - 1/100
=> A < 1
minh deo can ban k dau :((
\(a,\frac{1}{2}x+\frac{3}{5}(x-2)=3\)
\(\Rightarrow\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\Rightarrow\left[\frac{1}{2}+\frac{3}{5}\right]x=3+\frac{6}{5}\)
\(\Rightarrow\left[\frac{5}{10}+\frac{6}{10}\right]x=\frac{21}{5}\)
\(\Rightarrow\frac{11}{10}x=\frac{21}{5}\)
\(\Rightarrow x=\frac{21}{5}:\frac{11}{10}=\frac{21}{5}\cdot\frac{10}{11}=\frac{21}{1}\cdot\frac{2}{11}=\frac{42}{11}\)
Vậy x = 42/11
Ta xét A= \(\frac{1}{5^2}+\frac{1}{6^2}+..+\frac{1}{100^2}\)
\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}...+\frac{1}{100.101}\)
=> \(A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
=> \(A>\frac{1}{5}-\frac{1}{101}\)
=> \(A>\frac{96}{505}>\frac{96}{576}=\frac{1}{4}\)
Ta có : \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
=> \(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(A< \frac{1}{4}-\frac{1}{100}\)
=> \(A< \frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)
Đặt A = 1/5+1/6+1/7+...+1/17
Ta có :
1/5 + 1/6 + 1/7 + 1/8 + 1/9 + 1/10 < 1/5 + 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 6/5 (1)
1/11 + 1/12 + 1/13 + 1/14 + 1/15 + 1/16 + 1/17 < 1/11 + 1/11 + 1/11 + 1/11 +1/11 + 1/11 + 1/11 = 7/11 (2)
Từ (1) và (2) => :
A < 6/5 + 7/11 = 101/55 < 110/55 = 2
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}\)
\(=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\right)+\left(\frac{1}{11}+...+\frac{1}{17}\right)<\frac{1}{5}.6+\frac{1}{11}.7=\frac{6}{5}+\frac{7}{11}\)
\(=1\frac{46}{55}<2\)
\(\Rightarrowđpcm\)
\(A=\frac{3}{4}.\frac{8}{9}.........\frac{899}{900}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{29.31}{30^2}=\frac{1.2....29}{2.3....30}.\frac{3.4....31}{2.3....30}\)
\(=\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)
\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\right)+\left(\frac{1}{11}+...+\frac{1}{17}\right)\)
< 1/5 . 5 + 1/11.7 = 1+1/7 < 2
=>ĐPCM
\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{17}\right)\)
Ta có: \(\frac{1}{5}=\frac{1}{5};\frac{1}{6}< \frac{1}{5};\frac{1}{7}< \frac{1}{5};\frac{1}{8}< \frac{1}{5};\frac{1}{9}< \frac{1}{5}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}< \frac{1}{5}+\frac{1}{5}+...+\frac{1}{5}\left(5ps\right)=\frac{1}{5}\cdot5=1\left(1\right)\)
Lại có: \(\frac{1}{10}< \frac{1}{8};\frac{1}{11}< \frac{1}{8};...;\frac{1}{17}< \frac{1}{8}\)
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+...+\frac{1}{17}< \frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}\left(8ps\right)=\frac{1}{8}\cdot8=1\left(2\right)\)
Từ (1) và (2) => \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}< 1+1=2\)
P/s: k hỉu thì hỏi
Tại sao lại phải so sánh 5ps đầu vs 1/5 và các ps còn lại vs 1/8 mà ko phải là ps khác vậy?