đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

Ta có :

\(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

Lại có :

\(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)

Tu lam di

a) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}\)

\(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{44.45}\)

\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}\)

\(A< 1-\frac{1}{45}< 1\)

\(A< 1\)

\(\frac{x}{7}=\frac{x+1}{14}\Leftrightarrow14x=7x+7\Leftrightarrow7x=7\Leftrightarrow x=1\)

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{15}{4}+\frac{18}{8}\)

\(\Leftrightarrow1\le x\le6\Leftrightarrow x=1;2;3;4;5;6\)

\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)

\(\Leftrightarrow\frac{1}{2}-\frac{3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)

\(\Leftrightarrow0\le x\le5\Leftrightarrow x=0;1;2;3;4;5\)

\(\frac{x}{7}=\frac{x+1}{14}\)

=> \(\frac{x\cdot2}{7\cdot2}=\frac{x+1}{14}\)

=> \(2x=x+1\)

=> \(2x-x-1=0\)

=> \(1x-1=0\)

=> \(x=1\)

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\le x\le\frac{15}{4}+\frac{18}{8}\)

=> \(1\le x\le6\)

=> \(x=\left\{1;2;3;4;5;6\right\}\)

\(\frac{1}{2}+\frac{-3}{5}+\frac{1}{10}\le x\le\frac{8}{3}+\frac{14}{6}\)

=> \(0\le x\le5\)

=> \(x=\left\{0;1;2;3;4;5\right\}\)

Cảm ơn các bạn nhé !!!

giải tương tự như câu hôm qua mình giải

để chứng minh A < \(\frac{1}{10}\). Ta thấy \(A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)

\(\Rightarrow A^2< \left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\right)\)

\(=\frac{1.\left(3.5...99\right)}{2.4.6...100}.\frac{2.4.6...100}{\left(3.5.7...99\right).101}\)

\(=\frac{1}{101}< \frac{1}{10}\)

\(\Rightarrow A^2< \frac{1}{101}< \frac{1}{100}=\frac{1}{10^2}\Rightarrow A< \frac{1}{10}\)

để chứng minh A > \(\frac{1}{15}\). Ta thấy \(A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)

\(\Rightarrow A^2>\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\right).\left(\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\right)\)

\(=\frac{1.\left(3.5...99\right)}{\left(2.4.6...98\right).100}.\frac{1.\left(2.4...98\right)}{2.\left(3.5...99\right)}\)

\(=\frac{1}{100}.\frac{1}{2}=\frac{1}{200}\)

\(\Rightarrow A^2>\frac{1}{200}>\frac{1}{225}=\frac{1}{15^2}\Rightarrow A>\frac{1}{15}\)

CMR là gì vậy chị nếu em biết được thì có thể giải giùm chị em có công thức đây(lớp 5)