Question

Giải phương trình sau :

\((\dfrac{2}{x}+3)=(\dfrac{2}{x}+3)(x^2+1)\)

Answer 1

\(\left(\frac{2}{x}+3\right)=\left(\frac{2}{x}+3\right).\left(x^2+1\right)\) \(\left(ĐKXĐ:x\ne0\right).\)

\(\Leftrightarrow\left(\frac{2}{x}+3\right)-\left(\frac{2}{x}+3\right).\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(\frac{2}{x}+3\right).\left[1-\left(x^2+1\right)\right]=0\)

\(\Leftrightarrow\left(\frac{2}{x}+3\right).\left(1-x^2-1\right)=0\)

\(\Leftrightarrow\left(\frac{2}{x}+3\right).-x^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{2}{x}+3=0\\-x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{2}{x}=-3\\x=0\left(KTM\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\left(TM\right)\\x=0\left(KTM\right)\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-\frac{3}{2}\right\}.\)

Chúc bạn học tốt!

Answer 2

\(\left(\frac{2}{x}+3\right)=\left(\frac{2}{x}+3\right)\left(x^2+1\right)\\ \Leftrightarrow\left(\frac{2}{x}+3\right)-\left(\frac{2}{x}+3\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left(\frac{2}{x}+3\right)\cdot x^2=0\\ \Leftrightarrow\left[{}\begin{matrix}\frac{2}{x}+3=0\\x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{2}{x}=-3\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=0\end{matrix}\right.\)

Vậy \(S=\left\{\frac{-2}{3};0\right\}\)