Đáp án:
b) x=100
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{x + 1}}{{35}} + 1 + \dfrac{{x + 3}}{{33}} + 1 = \dfrac{{x + 5}}{{31}} + 1 + \dfrac{{x + 7}}{{29}} + 1\\
\to \dfrac{{x + 36}}{{35}} + \dfrac{{x + 36}}{{33}} = \dfrac{{x + 36}}{{31}} + \dfrac{{x + 36}}{{29}}\\
\to \left( {x + 36} \right)\left( {\dfrac{1}{{35}} + \dfrac{1}{{33}} - \dfrac{1}{{31}} - \dfrac{1}{{29}}} \right) = 0\\
\to x + 36 = 0\\
\to x = - 36\\
b)\dfrac{{x - 85}}{{15}} - 1 + \dfrac{{x - 74}}{{13}} - 2 + \dfrac{{x - 63}}{{11}} - 3 + \dfrac{{x - 64}}{9} - 4 = 0\\
\to \dfrac{{x - 100}}{{15}} + \dfrac{{x - 100}}{{13}} + \dfrac{{x - 100}}{{11}} + \dfrac{{x - 100}}{9} = 0\\
\to \left( {x - 100} \right)\left( {\dfrac{1}{{15}} + \dfrac{1}{{13}} + \dfrac{1}{{11}} + \dfrac{1}{9}} \right) = 0\\
\to x - 100 = 0\\
\to x = 100
\end{array}\)
