Question

a ) \(\dfrac{x+2}{x-2}\) - \(\dfrac{1}{x}\)= \(\dfrac{2}{x\left(x-2\right)}\)

b) \(\dfrac{x}{3}-\dfrac{x}{5}=4\)

 

 

Answer 1

\(a,\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\left(đkxđ:x\ne0;x\ne2\right)\)

\(\Leftrightarrow x\left(x+2\right)-x+2=2\)

\(\Leftrightarrow x^2+2x-x+2=2\)

\(\Leftrightarrow x^2+2x-x+2-2=0\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

Vậy \(S=\left\{-1\right\}\)

\(b,\dfrac{x}{3}-\dfrac{x}{5}=4\)

\(\Leftrightarrow5x-3x=60\)

\(\Leftrightarrow2x=60\)

\(\Leftrightarrow x=30\)

Vậy \(S=\left\{30\right\}\)