Question

Bài 2: Tìm x để các phân thức sau bằng 0
a, \(\dfrac{2x-1}{x^2-1}\)
b,\(\dfrac{\left(x+2\right)\left(x-3\right)}{2x-5}\)
c,\(\dfrac{x^2+x}{2x+1}\)
 

Answer 1

a)\(\dfrac{2x-1}{x^2-1}=0\)    (\(ĐKXĐ:x\ne\pm1\))

\(\Rightarrow2x-1=0\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

b)\(\dfrac{\left(x+2\right)\left(x-3\right)}{2x-5}=0\)      (\(ĐKXĐ:x\ne\dfrac{5}{2}\))

\(\Rightarrow\left(x+2\right)\left(x-3\right)=0\)

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

c)\(\dfrac{x^2+x}{2x+1}=0\)        (\(ĐKXĐ:x\ne-\dfrac{1}{2}\))

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

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