\(\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}=\frac{1}{8}\) (ĐKXĐ: x \(\ne\) -2; x \(\ne\) -3; x \(\ne\) -4; x \(\ne\) -5; x \(\ne\) -6)
\(\Leftrightarrow\) \(\frac{1}{x^2+2x+3x+6}+\frac{1}{x^2+3x+4x+12}+\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}=\frac{1}{8}\)
\(\Leftrightarrow\) \(\frac{1}{x\left(x+2\right)+3\left(x+2\right)}+\frac{1}{x\left(x+3\right)+4\left(x+3\right)}+\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}=\frac{1}{8}\)
\(\Leftrightarrow\) \(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\) \(\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\) \(\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\) \(\frac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\) \(\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\) \(\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\)
\(\Rightarrow\) (x + 2)(x + 6) = 32
\(\Leftrightarrow\) (x + 2)(x + 6) - 32 = 0
\(\Leftrightarrow\) x2 + 6x + 2x + 12 - 32 = 0
\(\Leftrightarrow\) x2 + 8x - 20 = 0
\(\Leftrightarrow\) x2 + 8x + 16 - 36 = 0
\(\Leftrightarrow\) (x + 4)2 - 36 = 0
\(\Leftrightarrow\) (x + 4 - 6)(x + 4 + 6) = 0
\(\Leftrightarrow\) (x - 2)(x + 10) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TMĐK\right)\\x=-10\left(TMĐK\right)\end{matrix}\right.\)
Vậy S = {2; -10}
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