Ta có :
\(\dfrac{\left(3x^2+7x-10\right)}{x}=0\), ĐKXĐ \(x\ne0\)
\(\Leftrightarrow3x^2+7x-10=0\)
\(\Leftrightarrow3x^2+7x-3-7=0\)
\(\Leftrightarrow3\left(x^2-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+10\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+10=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-10\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{10}{3}\left(TM\right)\\x=1\left(TM\right)\end{matrix}\right.\)
Vậy \(x=-\dfrac{10}{3}\) hoặc x = 1
\(ĐKXĐ:x\ne0\)
\(\dfrac{3x^2+7x-10}{x}=0\)
\(\Leftrightarrow3x^2+7x-10=0\)
\(\Leftrightarrow3x^2-3x+10x-10=0\)
\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{10}{3}\end{matrix}\right.\) ( t/m)
Vậy pt có tập nghiệm \(S=\left\{1;-\dfrac{10}{3}\right\}\)
em c.ơn trước
