Đáp án:
\(\begin{array}{l}
3)\,x = 2;x = \dfrac{2}{3}\\
4)\,a)\,x \ne \pm 2\\
b)M = \dfrac{{x + 2}}{{x - 2}}\\
c)\,M = \dfrac{1}{5}\\
d)\,x \in \left\{ {3;1;0;4;6} \right\}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2)\,a)\, \Leftrightarrow \left( {3x - 5y} \right)\left( {3x + 5y} \right):\left( {3x + 5y} \right)\\
= 3x - 5y\\
b)\, \Leftrightarrow {x^3} - {3^3} - 6{x^2} - 15x\\
= {x^3} - 6{x^2} - 15x - 27\\
c)\, \Leftrightarrow \dfrac{{3x + 2 - 3x + 2 + 3x - 6}}{{\left( {3x + 2} \right)\left( {3x - 2} \right)}}\\
= \dfrac{{3x - 2}}{{\left( {3x + 2} \right)\left( {3x - 2} \right)}} = \dfrac{1}{{3x + 2}}\\
3)\, \Leftrightarrow 3x\left( {x - 5} \right) - 2\left( {x - 5} \right) = 0\\
\Leftrightarrow \left( {x - 5} \right)\left( {3x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \dfrac{2}{3}
\end{array} \right.\\
4)\,a)\,DK:{x^2} - 4 \ne 0 \Leftrightarrow x \ne \pm 2\\
b)\,M = \dfrac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{x + 2}}{{x - 2}}\\
c)\,Thay\,x = - 3\,vao\,M\\
M = \dfrac{{ - 3 + 2}}{{ - 3 - 2}} = \dfrac{1}{5}\\
d)\,M = \dfrac{{x + 2}}{{x - 2}} = \dfrac{{x - 2 + 4}}{{x - 2}} = 1 + \dfrac{4}{{x - 2}}\\
M \in Z \Rightarrow \left( {x - 2} \right) \in U\left( 4 \right) = \left\{ { \pm 1; \pm 2; \pm 4} \right\}\\
\Rightarrow x \in \left\{ {3;1;0;4;6} \right\}
\end{array}\)
