\(\frac{x+2}{x-3}-\frac{3}{x-3}=\frac{1}{x}\)
\(\Leftrightarrow\frac{x\left(x+2\right)-3x}{x\left(x-3\right)}=\frac{x-3}{x\cdot\left(x-3\right)}\)
\(\Leftrightarrow x^2+2x-3x-x+3=0\)
\(\Leftrightarrow x^2-2x+3=0\)
\(\Delta=\left(-2\right)^2-4.3=-8< 0\)
Vậy phương trình vô nghiệm.
\(\frac{x+2}{x-3}-\frac{3}{x-3}=\frac{1}{x}\)
\(\Leftrightarrow\frac{\left(x+2\right)x}{\left(x-3\right)x}-\frac{3x}{\left(x-3\right).x}=\frac{\left(x-3\right)}{\left(x-3\right).x}\)
\(\Rightarrow x^2+2x-3x=x-3\)
\(\Leftrightarrow x^2-x-x-3=0\)
\(\Leftrightarrow x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow\left(x^2+x\right)-\left(3x+3\right)=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(\frac{x+2}{x-3}-\frac{3}{x-3}=\frac{1}{x}\) ( ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\) )
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-3\right)}-\frac{3x}{x\left(x-3\right)}=\frac{1\left(x-3\right)}{x\left(x-3\right)}\)
\(\Rightarrow x\left(x+2\right)-3x=x-3\)
\(\Leftrightarrow x^2+2x-3x=x-3\)
\(\Leftrightarrow x^2+2x-3x-x=-3\)
\(\Leftrightarrow x^2-2x=-3\)
\(\Leftrightarrow x^2-2x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow\left(x^2-3x\right)-\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tmĐKXĐ\right)\\x=3\left(kotmĐKXĐ\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình là \(x=1\) .
bổ sung thêm:
-điều kiện xác định \(x\ne3\) và \(x\ne0\)
Vậy nghiệm của phương trình là : x=-1
