a) ($\frac{2}{5}$ - x) : 1$\frac{1}{3}$+ $\frac{1}{2}$=-4 b) $\frac{x+1}{2014}$+ $\frac{x+2}{2015}$= $\frac{x+3}{2016}$+ $\frac{x+4}{2017}$ c) 3 - ( [x] ). $( 2.[x]+5 )^{2}$ =0 giúp mik với ! thks các bn
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Đáp án:
$\begin{array}{l}
a)\left( {\dfrac{2}{5} - x} \right):1\dfrac{1}{3} + \dfrac{1}{2} = - 4\\
\Leftrightarrow \left( {\dfrac{2}{5} - x} \right):\dfrac{4}{3} = - 4 - \dfrac{1}{2}\\
\Leftrightarrow \left( {\dfrac{2}{5} - x} \right):\dfrac{4}{3} = - \dfrac{9}{2}\\
\Leftrightarrow \dfrac{2}{5} - x = \dfrac{{ - 9}}{2}.\dfrac{4}{3}\\
\Leftrightarrow \dfrac{2}{5} - x = - 6\\
\Leftrightarrow x = \dfrac{2}{5} + 6\\
\Leftrightarrow x = \dfrac{{32}}{5}\\
Vậy\,x = \dfrac{{32}}{5}\\
b)\dfrac{{x + 1}}{{2014}} + \dfrac{{x + 2}}{{2015}} = \dfrac{{x + 3}}{{2016}} + \dfrac{{x + 4}}{{2017}}\\
\Leftrightarrow \dfrac{{x + 1}}{{2014}} - 1 + \dfrac{{x + 2}}{{2015}} - 1 = \dfrac{{x + 3}}{{2016}} - 1 + \dfrac{{x + 4}}{{2017}} - 1\\
\Leftrightarrow \dfrac{{x + 1 - 2014}}{{2014}} + \dfrac{{x + 2 - 2015}}{{2015}}\\
= \dfrac{{x + 3 - 2016}}{{2016}} + \dfrac{{x + 4 - 2017}}{{2017}}\\
\Leftrightarrow \dfrac{{x - 2013}}{{2014}} + \dfrac{{x - 2013}}{{2015}} = \dfrac{{x - 2013}}{{2016}} + \dfrac{{x - 2013}}{{2017}}\\
\Leftrightarrow \left( {x - 2013} \right).\left( {\dfrac{1}{{2014}} + \dfrac{1}{{2015}} - \dfrac{1}{{2016}} - \dfrac{1}{{2017}}} \right) = 0\\
\Leftrightarrow x = 2013\\
Vậy\,x = 2013\\
c)\left( {3 - \left| x \right|} \right){\left( {2\left| x \right| + 5} \right)^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left| x \right| = 3\\
2\left| x \right| = - 5\left( {ktm} \right)
\end{array} \right.\\
\Leftrightarrow x = 3;x = - 3\\
Vậy\,x = 3;x = - 3
\end{array}$