ĐKXĐ : \(x\ne\left\{2;3;4;5;6\right\}\)
\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4}{\left(x+2\right).\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right).\left(x+6\right)=32\)
\(\Leftrightarrow x^2+8x-20=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(\text{loại}\right)\\x=-10\end{matrix}\right.\Leftrightarrow x=-10\)
Vậy tập nghiệm phương trình S = {10}
a) x - 5 = 7 - x
<=> 2x = 12
<=> x = 6
Vậy tập nghiệm phương trình S = {6}
b) 3x - 15 = 2x(x - 5)
<=> 3(x - 5) = 2x(x - 5)
<=> (2x - 3)(x - 5) = 0
<=> \(\left[{}\begin{matrix}2x-3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=5\end{matrix}\right.\)
Tập nghiệm phương trình \(S=\left\{\dfrac{3}{2};5\right\}\)