a) \(\Rightarrow3x=-9\Rightarrow x=-3\)
b) \(\Rightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\dfrac{5\left(x+1\right)-15x-3\left(2x-1\right)}{15}=0\)
\(\Rightarrow5x+5-15x-6x+3=0\Rightarrow16x=8\Rightarrow x=\dfrac{1}{2}\)
d) \(x\ne0,x\ne2\)
\(\Rightarrow\dfrac{x\left(x+2\right)-\left(x-2\right)-2}{x\left(x-2\right)}=0\)
\(\Rightarrow x^2+2x-x+2-2=0\Rightarrow x^2+x=0\)
\(\Rightarrow x\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(a,\Leftrightarrow5x-2x=-7-2\)
\(\Leftrightarrow3x=-9\)
\(\Leftrightarrow x=-3\)
Vậy \(S=\left\{-3\right\}\)
\(b,\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Vậy \(S=\left\{2;-3\right\}\)
\(c,\Leftrightarrow\dfrac{15\left(x+1\right)-15x}{15}=\dfrac{3\left(2x-1\right)}{15}\)
\(\Leftrightarrow15x+15-15x-6x+3=0\left(quy\cdotđồng\cdot và\cdot khử\cdot mẫu\right)\)
\(\Leftrightarrow18-6x=0\)
\(\Leftrightarrow-6x=-18\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
Câu (d) áp dụng tương tự như (c) , quy đồng và khử mẫu
