Đáp án:
f) x=2018
Giải thích các bước giải:
\(\begin{array}{l}
b)\left( {x + 1} \right)\left( {x + 2} \right) - \left( {2 - x} \right)\left( {x + 2} \right) = 0\\
\to \left( {x + 2} \right)\left( {x + 1 - 2 + x} \right) = 0\\
\to \left[ \begin{array}{l}
x + 2 = 0\\
2x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
x = \dfrac{1}{2}
\end{array} \right.\\
c)DK:x \ne \pm 5\\
\dfrac{{{x^2} + 10x + 25 - {x^2} + 10x - 25}}{{\left( {x - 5} \right)\left( {x + 5} \right)}} = \dfrac{{20}}{{\left( {x - 5} \right)\left( {x + 5} \right)}}\\
\to 20x = 20\\
\to x = 1\\
d)\dfrac{{2 - x}}{{2016}} + 1 = \dfrac{{1 - x}}{{2017}} + 1 - \dfrac{x}{{2018}} + 1\\
\to \dfrac{{2018 - x}}{{2016}} = \dfrac{{2018 - x}}{{2017}} + \dfrac{{2018 - x}}{{2018}}\\
\to \dfrac{{2018 - x}}{{2016}} - \dfrac{{2018 - x}}{{2017}} - \dfrac{{2018 - x}}{{2018}} = 0\\
\to \left( {2018 - x} \right)\left( {\dfrac{1}{{2016}} - \dfrac{1}{{2017}} - \dfrac{1}{{2018}}} \right) = 0\\
\to 2018 - x = 0\\
\to x = 2018\\
e){\left( {x + 3} \right)^2} = 144\\
\to \left| {x + 3} \right| = 12\\
\to \left[ \begin{array}{l}
x + 3 = 12\\
x + 3 = - 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = - 15
\end{array} \right.\\
f)\dfrac{{x - 19}}{{1999}} - 1 + \dfrac{{x - 23}}{{1995}} - 1 + \dfrac{{x + 82}}{{700}} - 3 = 0\\
\to \dfrac{{x - 2018}}{{1999}} + \dfrac{{x - 2018}}{{1995}} + \dfrac{{x - 2018}}{{700}} = 0\\
\to \left( {x - 2018} \right)\left( {\dfrac{1}{{1999}} + \dfrac{1}{{1995}} + \dfrac{1}{{700}}} \right) = 0\\
\to x - 2018 = 0\\
\to x = 2018
\end{array}\)
