\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
\(\Leftrightarrow\) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}+\frac{4\left(x+10\right)\left(2-x\right)}{12}=0\)
\(\Leftrightarrow\) (x + 10)(x + 4) - 3(x + 4)(2 - x) + 4(x + 10)(2 - x) = 0
\(\Leftrightarrow\) x2 + 14x + 40 + 3x2 + 6x - 24 - 4x2 - 32x + 80 = 0
\(\Leftrightarrow\) -12x + 96 = 0
\(\Leftrightarrow\) -12(x - 8) = 0
\(\Leftrightarrow\) x - 8 = 0
\(\Leftrightarrow\) x = 8
Vậy S = {8}
Chúc bn học tốt!!
