Ta có:
1x2+9x+20+1x2+11x+30+1x2+13x+42=1181x2+9x+20+1x2+11x+30+1x2+13x+42=118 (DK:x≠−4;x≠−5;x≠−6;x≠−7)(DK:x≠−4;x≠−5;x≠−6;x≠−7)
\Leftrightarrow1(x+4)(x+5)+1(x+5)(x+6)+1(x+6)(x+7)=1181(x+4)(x+5)+1(x+5)(x+6)+1(x+6)(x+7)=118
\Leftrightarrow1x+4−1x+5+1x+5−1x+6+1x+6−1x+7=1181x+4−1x+5+1x+5−1x+6+1x+6−1x+7=118
\Leftrightarrow1x+4−1x+7=1181x+4−1x+7=118
\Leftrightarrow3x2+11x+28=1183x2+11x+28=118
\Rightarrow x2+11x+28=54x2+11x+28=54
\Leftrightarrowx2+11x−26=0x2+11x−26=0
\Leftrightarrow (x−2)(x+13)=0(x−2)(x+13)=0
\Leftrightarrow\left[\begin{x-2=0}\\{x+13 = 0}\left[\begin{x-2=0}\\{x+13 = 0}
\Leftrightarrow\left[\begin{x=2}(t/m)\\{x= -13} (t/m)\left[\begin{x=2}(t/m)\\{x= -13} (t/m)
Vậy pt có tập nghiệm là S={2,-13}

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
eq4 ; x
eq5 ; x
eq 6 ; x
eq7 )[/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}

\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{ x^2+13x+42}=\frac{1}{18}
\Leftrightarrow\frac {1}{x^2+5x+4x+20}+\frac{1}{x^2+6x+5x+30}+\frac{1}{ x^2+7x+6x+42}=\frac{1}{18}
\Leftrightarrow\frac {1}{x(x+5)+4(x+5)}+\frac{1}{x(x+6)+5(x+6)}+\frac{1 }{x(x+7)+6(x+7)}=\frac{1}{18}
\Leftrightarrow\frac {1}{(x+4)(x+5)}+\frac{1}{(x+5)(x+6)}+\frac{1}{(x+6 )(x+7)}=\frac{1}{18}
ĐKXĐ: \frac{x}{-4}
eq, \frac{x}{-5}
eq, \frac{x}{-6}
eq và \frac{x}{-7}
eq
\frac{1}{(x+4)(x+5)}+\frac{1}{(x+5)(x+6)}+\frac{1} {(x+6)(x+7)}=\frac{1}{18}
\Leftrightarrow \frac{x+5-x-4}{(x+4)(x+5)}+\frac{x+6-x-5}{(x+5)(x+6)}+\frac{x+7-x-6}{(x+6)(x+7)}={1}{18}
\Leftrightarrow \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}
\Leftrightarrow \frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}

Chú ý tớ gõ công thức Toán, Vật lí, Hóa học