Giải phương trình:
1/x^2+9x+20 + 1/x^2+11x+30 +1/x^2+13x+42 = 1/18
Ta có:
[TEX]\frac{1}{x^2+9x+20} + \frac{1}{x^2+11x+30} + \frac{1}{x^2+13x+42} = \frac{1}{18}[/TEX] [TEX](DK: x\neq-4 ; x\neq-5 ; x\neq -6 ; x\neq-7 )[/TEX]
\Leftrightarrow[TEX]\frac{1}{(x+4)(x+5)} + \frac{1}{(x+5)(x+6)} + \frac{1}{(x+6)(x+7)} = \frac{1}{18}[/TEX]
\Leftrightarrow[TEX]\frac{1}{x+4} - \frac{1}{x+5} +\frac{1}{x+5} - \frac{1}{x+6} + \frac{1}{x+6} - \frac{1}{x+7} = \frac{1}{18} [/TEX]
\Leftrightarrow[TEX]\frac{1}{x+4} - \frac{1}{x+7}= \frac {1}{18} [/TEX]
\Leftrightarrow[TEX]\frac{3}{x^2+11x+28}=\frac{1}{18}[/TEX]
\Rightarrow [TEX]x^2+ 11x + 28 = 54[/TEX]
\Leftrightarrow[TEX]x^2+11x-26=0[/TEX]
\Leftrightarrow [TEX](x-2)(x+13)=0[/TEX]
\Leftrightarrow[tex]\left[\begin{x-2=0}\\{x+13 = 0}[/tex]
\Leftrightarrow[tex]\left[\begin{x=2}(t/m)\\{x= -13} (t/m)[/tex]
Vậy pt có tập nghiệm là S={2,-13}