Question

a)\(\frac{x-2}{2}-\frac{x-2}{6}+\frac{2-x}{3}=0\)

b)\(\left(x-2012\right)\left(3x+8\right)=0\)

c)\(\frac{2}{x+3}-\frac{3}{x-2}=\frac{x+4}{\left(x+3\right)\left(x-2\right)}\)

Answer 1

a, \(\frac{x-2}{2}-\frac{x-2}{6}+\frac{2-x}{3}=0\)

\(\Leftrightarrow\) \(\frac{x-2}{2}-\frac{x-2}{6}-\frac{x-2}{3}=0\)

\(\Leftrightarrow\) (x - 2)(\(\frac{1}{2}-\frac{1}{6}-\frac{1}{3}\)) = 0

\(\Leftrightarrow\) x - 2 = 0

\(\Leftrightarrow\) x = 2

Vậy S = {2}

b, (x - 2012)(3x + 8) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-2012=0\\3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2012\\x=\frac{-8}{3}\end{matrix}\right.\)

Vậy S = {2012; \(\frac{-8}{3}\)}

c, \(\frac{2}{x+3}-\frac{3}{x-2}=\frac{x+4}{\left(x+3\right)\left(x-2\right)}\) (ĐKXĐ: x \(\ne\) -3; x \(\ne\) 2)

\(\Leftrightarrow\) \(\frac{2\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{3\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x+4}{\left(x+3\right)\left(x-2\right)}\)

\(\Rightarrow\) 2(x - 2) - 3(x + 3) = x + 4

\(\Leftrightarrow\) 2x - 4 - 3x - 9 - x - 4 = 0

\(\Leftrightarrow\) -2x - 17 = 0

\(\Leftrightarrow\) x = \(\frac{-17}{2}\)

Vậy S = {\(\frac{-17}{2}\)}

Chúc bn học tốt!!