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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 \(x = 5\)
Hãy giúp mọi người biết câu trả lời này thế nào?
Giải thích các bước giải:
Ta có:
(11.51+12.52+13.53+....+110.60)x=(11.11+12.12+...+150.60)A=11.51+12.52+13.53+....+110.60⇔50A=501.51+502.52+503.53+....+5010.60⇔50A=51−11.51+52−22.52+53−33.53+....+60−1010.60⇔50A=1−151+12−152+13−153+....+110−160⇔50A=(1+12+13+....+110)−(151+152+....+160)B=11.11+12.12+13.13+.....+150.60⇔10B=101.11+102.12+103.13+....+1050.60⇔10B=11−11.11+12−22.12+13−33.13+....+60−5050.60⇔10B=1−111+12−112+13−113+....+150−160⇔10B=(1+12+13+....+150)−(111+112+113+....+160)⇔10B=(1+12+13+...+110)−(151+152+153+.....+160)⇒50A=10B⇔BA=5x=BA=5(11.51+12.52+13.53+....+110.60)x=(11.11+12.12+...+150.60)A=11.51+12.52+13.53+....+110.60⇔50A=501.51+502.52+503.53+....+5010.60⇔50A=51−11.51+52−22.52+53−33.53+....+60−1010.60⇔50A=1−151+12−152+13−153+....+110−160⇔50A=(1+12+13+....+110)−(151+152+....+160)B=11.11+12.12+13.13+.....+150.60⇔10B=101.11+102.12+103.13+....+1050.60⇔10B=11−11.11+12−22.12+13−33.13+....+60−5050.60⇔10B=1−111+12−112+13−113+....+150−160⇔10B=(1+12+13+....+150)−(111+112+113+....+160)⇔10B=(1+12+13+...+110)−(151+152+153+.....+160)⇒50A=10B⇔BA=5x=BA=5
Vậy x=5x=5
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