Đáp án:
\[x = 5\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left( {\frac{1}{{1.51}} + \frac{1}{{2.52}} + \frac{1}{{3.53}} + .... + \frac{1}{{10.60}}} \right)x = \left( {\frac{1}{{1.11}} + \frac{1}{{2.12}} + ... + \frac{1}{{50.60}}} \right)\\
A = \frac{1}{{1.51}} + \frac{1}{{2.52}} + \frac{1}{{3.53}} + .... + \frac{1}{{10.60}}\\
\Leftrightarrow 50A = \frac{{50}}{{1.51}} + \frac{{50}}{{2.52}} + \frac{{50}}{{3.53}} + .... + \frac{{50}}{{10.60}}\\
\Leftrightarrow 50A = \frac{{51 - 1}}{{1.51}} + \frac{{52 - 2}}{{2.52}} + \frac{{53 - 3}}{{3.53}} + .... + \frac{{60 - 10}}{{10.60}}\\
\Leftrightarrow 50A = 1 - \frac{1}{{51}} + \frac{1}{2} - \frac{1}{{52}} + \frac{1}{3} - \frac{1}{{53}} + .... + \frac{1}{{10}} - \frac{1}{{60}}\\
\Leftrightarrow 50A = \left( {1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{{10}}} \right) - \left( {\frac{1}{{51}} + \frac{1}{{52}} + .... + \frac{1}{{60}}} \right)\\
B = \frac{1}{{1.11}} + \frac{1}{{2.12}} + \frac{1}{{3.13}} + ..... + \frac{1}{{50.60}}\\
\Leftrightarrow 10B = \frac{{10}}{{1.11}} + \frac{{10}}{{2.12}} + \frac{{10}}{{3.13}} + .... + \frac{{10}}{{50.60}}\\
\Leftrightarrow 10B = \frac{{11 - 1}}{{1.11}} + \frac{{12 - 2}}{{2.12}} + \frac{{13 - 3}}{{3.13}} + .... + \frac{{60 - 50}}{{50.60}}\\
\Leftrightarrow 10B = 1 - \frac{1}{{11}} + \frac{1}{2} - \frac{1}{{12}} + \frac{1}{3} - \frac{1}{{13}} + .... + \frac{1}{{50}} - \frac{1}{{60}}\\
\Leftrightarrow 10B = \left( {1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{{50}}} \right) - \left( {\frac{1}{{11}} + \frac{1}{{12}} + \frac{1}{{13}} + .... + \frac{1}{{60}}} \right)\\
\Leftrightarrow 10B = \left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{10}}} \right) - \left( {\frac{1}{{51}} + \frac{1}{{52}} + \frac{1}{{53}} + ..... + \frac{1}{{60}}} \right)\\
\Rightarrow 50A = 10B\\
\Leftrightarrow \frac{B}{A} = 5\\
x = \frac{B}{A} = 5
\end{array}\)
Vậy \(x = 5\)
