Bài 3: Tìm x:
g. 3x – 2|x + 1| = 5
h. x – 2 / x + 3 = 2 / 3
i. x – 1 / 24 = 6 / x – 1
k. x-2 / 2016 + x-3 / 2015 + x-4 / 2014 + x-5 / 2013 = 4
m. 5 √x – 3 – 4 =16
n. ( x – 1) ² – 3 (x – 1) = 0
Bài 3: Tìm x:
g. 3x – 2|x + 1| = 5
h. x – 2 / x + 3 = 2 / 3
i. x – 1 / 24 = 6 / x – 1
k. x-2 / 2016 + x-3 / 2015 + x-4 / 2014 + x-5 / 2013 = 4
m. 5 √x – 3 – 4 =16
n. ( x – 1) ² – 3 (x – 1) = 0
Đáp án:
$\begin{array}{l}
g)3x – 2\left| {x + 1} \right| = 5\\
\Rightarrow 2\left| {x + 1} \right| = 3x – 5\\
\Rightarrow \left[ \begin{array}{l}
2x + 2 = 3x – 5\\
2x + 2 = – 3x + 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 7\\
x = \dfrac{3}{5}
\end{array} \right.\\
h)\dfrac{{x – 2}}{{x + 3}} = \dfrac{2}{3}\left( {dk:x \ne – 3} \right)\\
\Rightarrow 3\left( {x – 2} \right) = 2\left( {x + 3} \right)\\
\Rightarrow 3x – 6 = 2x + 6\\
\Rightarrow x = 12\left( {tm} \right)\\
i)\dfrac{{x – 1}}{{24}} = \dfrac{6}{{x – 1}}\left( {dk:x \ne 1} \right)\\
\Rightarrow {\left( {x – 1} \right)^2} = 24.6 = 144\\
\Rightarrow \left[ \begin{array}{l}
x – 1 = 12\\
x – 1 = – 12
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 13\\
x = – 11
\end{array} \right.\left( {tm} \right)\\
k)\dfrac{{x – 2}}{{2016}} + \dfrac{{x – 3}}{{2015}} + \dfrac{{x – 4}}{{2014}} + \dfrac{{x – 5}}{{2013}} = 4\\
\Rightarrow \dfrac{{x – 2}}{{2016}} – 1 + \dfrac{{x – 3}}{{2015}} – 1 + \dfrac{{x – 4}}{{2014}} – 1 + \dfrac{{x – 5}}{{2013}} – 1 = 0\\
\Rightarrow \dfrac{{x – 2 – 2016}}{{2016}} + \dfrac{{x – 3 – 2015}}{{2015}} + \\
\dfrac{{x – 4 – 2014}}{{2014}} + \dfrac{{x – 5 – 2013}}{{2013}} = 0\\
\Rightarrow \dfrac{{x – 2018}}{{2016}} + \dfrac{{x – 2018}}{{2015}} + \dfrac{{x – 2018}}{{2014}} + \dfrac{{x – 2018}}{{2013}} = 0\\
\Rightarrow \left( {x – 2018} \right)\left( {\dfrac{1}{{2016}} + \dfrac{1}{{2015}} + \dfrac{1}{{2014}} + \dfrac{1}{{2013}}} \right) = 0\\
\Rightarrow x – 2018 = 0\\
\Rightarrow x = 2018\\
m)5\sqrt {x – 3} – 4 = 16\left( {dkxd:x \ge 3} \right)\\
\Rightarrow 5\sqrt {x – 3} = 16 + 4 = 20\\
\Rightarrow \sqrt {x – 3} = 4\\
\Rightarrow x – 3 = 16\\
\Rightarrow x = 19\left( {tm} \right)\\
n){\left( {x – 1} \right)^2} – 3\left( {x – 1} \right) = 0\\
\Rightarrow \left( {x – 1} \right)\left( {x – 1 – 3} \right) = 0\\
\Rightarrow \left( {x – 1} \right)\left( {x – 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x – 1 = 0\\
x – 4 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = 4
\end{array} \right.
\end{array}$