Question

giúp mình ý b thôi

Answer 1

\(\dfrac{3a+2}{3a+4}+\dfrac{a-2}{a+4}=2\)

ĐKXĐ: \(a\ne-4;a\ne-\dfrac{4}{3}\)

\(\Leftrightarrow\dfrac{\left(3a+2\right)\left(a+4\right)}{\left(3a+4\right)\left(a+4\right)}+\dfrac{\left(3a+4\right)\left(a-2\right)}{\left(3a+4\right)\left(a+4\right)}=\dfrac{2.\left(3a+4\right)\left(a+4\right)}{\left(3a+4\right)\left(a+4\right)}\)

\(\Rightarrow3a^2+14a+8+3a^2-2a-8=2.\left(3a^2+16a+16\right)\)

\(\Leftrightarrow6a^2+12a=6a^2+32a+32\\ \Leftrightarrow6a^2-6a^2+12a-32a-32=0\\ \Leftrightarrow-20a-32=0\\ \Leftrightarrow-20a=32\\ \Leftrightarrow a=-\dfrac{32}{20}\)

\(\Leftrightarrow a=-\dfrac{8}{5}\) ( nhận )

Vậy S = \(\left\{-\dfrac{8}{5}\right\}\)