Đáp án:
e) x=3
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{2 - x}}{{2001}} - 1 = \dfrac{{1 - x}}{{2002}} - \dfrac{x}{{2003}}\\
\to \dfrac{{2 - x}}{{2001}} + 1 = \dfrac{{1 - x}}{{2002}} + 1 - \dfrac{x}{{2003}} + 1\\
\to \dfrac{{2003 - x}}{{2001}} = \dfrac{{2003 - x}}{{2002}} + \dfrac{{ - x + 2003}}{{2003}}\\
\to \left( {2003 - x} \right)\left( {\dfrac{1}{{2001}} - \dfrac{1}{{2002}} - \dfrac{1}{{2003}}} \right) = 0\\
\to 2003 - x = 0\\
\to x = 2003\\
b)2x + 4\left( {36 - x} \right) = 100\\
\to 2x + 144 - 4x = 100\\
\to 2x = 44\\
\to x = 22\\
c)4x + 2\left( {36 - x} \right) = 100\\
\to 4x + 72 - 2x = 100\\
\to 2x = 28\\
\to x = 14\\
d)DK:x \ne - 5\\
\dfrac{{x + 2}}{{x + 5}} = \dfrac{1}{2}\\
\to 2x + 4 = x + 5\\
\to x = 1\\
e)DK:x \ne \left\{ { - 1;0} \right\}\\
\dfrac{{x - 1}}{x} - \dfrac{{2x - 1}}{{{x^2} + x}} = \dfrac{1}{{x + 1}}\\
\dfrac{{{x^2} - 1 - 2x + 1 - x}}{{x\left( {x + 1} \right)}} = 0\\
\to \dfrac{{{x^2} - 3x}}{{x\left( {x + 1} \right)}} = 0\\
\to \dfrac{{x\left( {x - 3} \right)}}{{x\left( {x + 1} \right)}} = 0\\
\to \dfrac{{x - 3}}{{x + 1}} = 0\\
\to x - 3 = 0\\
\to x = 3
\end{array}\)
