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

Cảm ơn bạn rất nhiều!

Ta có:

\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(=1-\frac{1}{51}=\frac{50}{51}\)

\(\Rightarrow A=\frac{50}{51}:2=\frac{25}{51}\)

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)

Ta  có: \(\frac{1}{2^2}>\frac{1}{100}\)

\(\frac{1}{4^2}>\frac{1}{100}\)

...

\(\frac{1}{100^2}< \frac{1}{100}\)

=> \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)

Đặt  \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{100^2}\)

\(\Rightarrow A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)

Ta có:

\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1-\frac{1}{50}< 1\)

\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+1=2\Rightarrow A< \frac{1}{2^2}.2\Rightarrow A< \frac{1}{2}\left(đpcm\right)\)

\(A=1+\frac{2^2}{3^2}+\frac{2^2}{5^2}+\frac{2^2}{7^2}+...+\frac{2^2}{2009^2}\)

\(A=1+2^2\left(\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+..+\frac{1}{2009^2}\right)\)

Ta có: \(\frac{1}{3^2}< \frac{1}{1.3};\frac{1}{5^2}< \frac{1}{3.5};\frac{1}{7^2}< \frac{1}{5.7};...;\frac{1}{2009^2}< \frac{1}{2007.2009}\)

\(\Rightarrow A< 1+4\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{2007.2009}\right)\)

\(=1+4\cdot\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2007}-\frac{1}{2009}\right)\)

\(=1+2\left(1-\frac{1}{2009}\right)=3-\frac{2}{2009}< 3\)

\(\Rightarrow A< 3\)

chúc bạn may mắn lần sau !!!! (^^!)

hỏi chị google ấy

A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

= \(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)

=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)

=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)

=\(\frac{-5321}{4805}+\frac{30}{31}\)

=\(\frac{-671}{4805}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)=\frac{7}{15}\)

Ta có: \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}\right)\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\)

\(\Leftrightarrow\frac{7}{15}\left(x-1\right)=\frac{3}{5}x-\frac{7}{15}\Leftrightarrow\frac{7}{15}x=\frac{3}{5}x\)

\(\Leftrightarrow\frac{2}{15}x=0\Leftrightarrow x=0\)

Tập nghiệm: \(S=\left\{0\right\}\)

a, Ta có : \(\frac{x+1}{2}+\frac{x-2}{4}=1-\frac{2\left(x-1\right)}{3}\)

=> \(\frac{6\left(x+1\right)}{12}+\frac{3\left(x-2\right)}{12}=\frac{12}{12}-\frac{8\left(x-1\right)}{12}\)

=> \(6\left(x+1\right)+3\left(x-2\right)=12-8\left(x-1\right)\)

=> \(6x+6+3x-6=12-8x+8\)

=> \(17x=20\)

=> \(x=\frac{20}{17}\)

b, Ta có : \(\frac{5x-1}{6}+x=\frac{6-x}{4}\)

=> \(\frac{5x-1+6x}{6}=\frac{6-x}{4}\)

=> \(4\left(11x-1\right)=6\left(6-x\right)\)

=> \(44x-4-36+6x=0\)

=> \(\)\(50x=40\)

=> \(x=\frac{4}{5}\)

c, Ta có : \(\frac{5\left(1-2x\right)}{3}+\frac{x}{2}=\frac{3\left(x-5\right)}{4}-2\)

=> \(\frac{20\left(1-2x\right)}{12}+\frac{6x}{12}=\frac{9\left(x-5\right)}{12}-\frac{24}{12}\)

=> \(20\left(1-2x\right)+6x=9\left(x-5\right)-24\)

=> \(20-40x+6x-9x+45+24=0\)

=> \(43x=89\)

=> \(x=\frac{89}{43}\)