Đáp án:
c) x=-2005
Giải thích các bước giải:
\(\begin{array}{l}
a)\left( {x - 23} \right)\left( {\dfrac{1}{{24}} + \dfrac{1}{{25}} - \dfrac{1}{{26}} - \dfrac{1}{{27}}} \right) = 0\\
\to x - 23 = 0\\
\to x = 23\\
b)\dfrac{{x + 2 + 98}}{{98}} + \dfrac{{x + 3 + 97}}{{97}} = \dfrac{{x + 4 + 96}}{{96}} + \dfrac{{x + 5 + 95}}{{95}}\\
\to \dfrac{{x + 100}}{{98}} + \dfrac{{x + 100}}{{97}} - \dfrac{{x + 100}}{{96}} - \dfrac{{x + 100}}{{95}} = 0\\
\to \left( {x + 100} \right)\left( {\dfrac{1}{{98}} + \dfrac{1}{{97}} - \dfrac{1}{{96}} - \dfrac{1}{{95}}} \right) = 0\\
\to x + 100 = 0\\
\to x = - 100\\
c)\dfrac{{x + 1}}{{2004}} + 1 + \dfrac{{x + 2}}{{2003}} + 1 = \dfrac{{x + 3}}{{2002}} + 1 + \dfrac{{x + 4}}{{2001}} + 1\\
\to \dfrac{{x + 2005}}{{2004}} + \dfrac{{x + 2005}}{{2003}} = \dfrac{{x + 2005}}{{2002}} + \dfrac{{x + 2005}}{{2001}}\\
\to \left( {x + 2005} \right)\left( {\dfrac{1}{{2004}} + \dfrac{1}{{2003}} - \dfrac{1}{{2002}} - \dfrac{1}{{2001}}} \right) = 0\\
\to x + 2005 = 0\\
\to x = - 2005\\
d)\dfrac{{201 - x}}{{99}} + \dfrac{{203 - x}}{{97}} + \dfrac{{205 - x}}{{95}} + 3 = 0\\
\to \dfrac{{201 - x}}{{99}} + 1 + \dfrac{{203 - x}}{{97}} + 1 + \dfrac{{205 - x}}{{95}} + 1 = 0\\
\to \dfrac{{300 - x}}{{99}} + \dfrac{{300 - x}}{{97}} + \dfrac{{300 - x}}{{95}} = 0\\
\to \left( {300 - x} \right)\left( {\dfrac{1}{{99}} + \dfrac{1}{{97}} + \dfrac{1}{{95}}} \right) = 0\\
\to 300 - x = 0\\
\to x = 300
\end{array}\)
