\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)ĐKXĐ: x \(\ne\)0,x\(\ne\)-10
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\Leftrightarrow\frac{12\left(x+10\right)}{12x\left(x+10\right)}+\frac{12x}{12x\left(x+10\right)}=\frac{x\left(x+10\right)}{12x\left(x+10\right)}\Rightarrow12x+120+12x=x^2+10x\Leftrightarrow x^2+10x-24x-120=0\Leftrightarrow x^2-14x-120=0\Leftrightarrow x^2-20x+6x-120=0\Leftrightarrow x\left(x-20\right)+6\left(x-20\right)\Leftrightarrow\left(x+6\right)\left(x-20\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=20\end{matrix}\right.\)Vậy S=\(\left\{-6;20\right\}\)
Theo mik thì chắc đảo hết phân số lại
+) Đkxđ :
x \(\ne\)0
x + 10 \(\ne\)0 \(\rightarrow\)x \(\ne\)-10
\(\Rightarrow\)12(x + 10) + 12 = x + 10
\(\Leftrightarrow\)12x + 120 + 12 = x + 10
\(\Leftrightarrow\) 12x + 132 = x + 10
\(\Leftrightarrow\) 12x - x = 10 - 132
\(\Leftrightarrow\) 11x = -122
\(\Leftrightarrow\) x = -11 ( TMĐK )
Vậy S = \(\left\{-11\right\}\)