c)$\dfrac{x}{x-2}+\dfrac{1}{x+3}=\dfrac{x^2+2}{(x-2)(x+3)}$
ĐKXĐ: `x ne -3` và `x ne2`
⇔$\dfrac{x(x+3)}{(x-2)(x+3)}+\dfrac{x-2}{(x+3)(x-2)}=\dfrac{x^2+2}{(x-2)(x+3)}$
`⇒x^2+3x+x-2=x^2+2`
`⇔4x=4`
`⇔x=1(N)`
Vậy `S={1}`
`d)|x+2|=3x+4`
ĐK:`3x+4≥0`
`⇔3x≥-4`
`⇔x≥-4/3`
⇔\(\left[ \begin{array}{l}x+2=3x+4\\x+2=-3x-4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x-3x=4-2\\x+3x=-4-2\end{array} \right.\)
⇔\(\left[ \begin{array}{l}-2x=2\\4x=-6\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-1(N)\\x=-3/2(L)\end{array} \right.\)
Vậy `S={-1}`
