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Ta có:
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(=1-\frac{1}{51}=\frac{50}{51}\)
\(\Rightarrow A=\frac{50}{51}:2=\frac{25}{51}\)
Ta có: \(1=\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\)
Vì a,b,c là số nguyên dương nên:
Ta có: \(\frac{a}{a+b}>\frac{a}{a+b+c}\)
\(\frac{b}{b+c}>\frac{b}{a+b+c}\)
\(\frac{c}{c+a}>\frac{c}{a+b+c}\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)
đpcm
mk xin lỗi các bạn nhé
1/tìm p/s ... 8 lần p/số a/b
nhé hihi
I'm sorry
Giải thích thêm: ta thấy \(\frac{1}{2^2}>\frac{1}{100}\),...,\(\frac{1}{10^2}=\frac{1}{100}\)=> từ \(\frac{1}{2^2}\)đến \(\frac{1}{10^2}\)có 5 cặp
\(\frac{1}{12^2}< \frac{1}{100}\),...,\(\frac{1}{100^2}< \frac{1}{100}\)=> từ \(\frac{1}{12^2}\)đến \(\frac{1}{100^2}\)có 45 cặp
=> 45>5 => tổng < 1/2 (kết hợp với cái kia nx thì bn mới hiểu)
\(A=1+\frac{2^2}{3^2}+\frac{2^2}{5^2}+\frac{2^2}{7^2}+...+\frac{2^2}{2009^2}\)
\(A=1+2^2\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+..+\frac{1}{2009^2}\right)\)
Ta có: \(\frac{1}{3^2}< \frac{1}{1.3};\frac{1}{5^2}< \frac{1}{3.5};\frac{1}{7^2}< \frac{1}{5.7};...;\frac{1}{2009^2}< \frac{1}{2007.2009}\)
\(\Rightarrow A< 1+4\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{2007.2009}\right)\)
\(=1+4\cdot\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2009}\right)\)
\(=1+2\left(1-\frac{1}{2009}\right)=3-\frac{2}{2009}< 3\)
\(\Rightarrow A< 3\)
1) K = D. 10 000 + Q
=> K-Q = D.10 000
=> 2015(K-Q) + 2016D = 2015.D.10 000 + 2016D =20152016.D
Vậy 2015(K-Q) + 2016D chia cho D = 20152016D:D = 20152016
2) \(A=\frac{\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right)}{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}=\)
\(A=\frac{\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}\right)}{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}=\)
\(=\frac{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}=1\)
A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
= \(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)
=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)
=\(\frac{-5321}{4805}+\frac{30}{31}\)
=\(\frac{-671}{4805}\)
a/b= (1+1/6) + (1/2+1/5) + (1/3+1/4)
a/b= 7/6 + 7/10 + 7/12
a/b= 7(1/6+1/10+1/12)
Vì 6x10x12 khong la boi so cua 7 => a/b chia het cho 7 <=> a chia het cho 7 (dpcm)
Bạn ơi cho mình hỏi dpcm là gì vậy?