1) điều kiện xác định : \(x\notin\left\{-1;-3;-5;-7\right\}\)

ta có : \(\dfrac{1}{x^2+4x+3}+\dfrac{1}{x^2+8x+15}+\dfrac{1}{x^2+12x+35}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{x^2+12x+35+x^2+8x+7+x^2+4x+3}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{3x^2+24x+45}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}=\dfrac{1}{9}\)

\(\Leftrightarrow9\left(3x^2+24x+45\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)

\(\Leftrightarrow27\left(x^2+8x+15\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)

\(\Leftrightarrow27\left(x+3\right)\left(x+5\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)\)

\(\Leftrightarrow27=\left(x+1\right)\left(x+7\right)\) ( vì điều kiện xác định )

\(\Leftrightarrow27=x^2+8x+7\Leftrightarrow x^2+8x-20=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\left(tmđk\right)\)

vậy \(x=2\) hoặc \(x=-10\)

\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{x-6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow32=\left(x-2\right)\left(x-6\right)\)

\(\Leftrightarrow32=x^2-8x+12\)

\(\Leftrightarrow x^2+8x-20=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)

a.

\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow x\left(x+1\right).\left(x-1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt \(a=x^2+x-1\) , ta có pt:

\(\left(a+1\right)\left(a-1\right)-24=0\)

\(\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\)

\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)

*Với a = 5 ta được:

\(x^2+x-1=5\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow x^2+3x-2x-6=0\)

\(\Leftrightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)

\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

*Với a = -5 ta được:

\(x^2+x-1=-5\)

\(\Leftrightarrow x^2+x+4=0\)

\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)

Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)

c)(ĐKXĐ: x khác 30;29)

\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)

\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)

\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)

\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)

\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)

\(\Leftrightarrow1741-59x=0\)

\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)

Vậy S={0;\(\dfrac{1741}{59}\);59}

b)(ĐKXĐ:x khác 2;3;4;5;6)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-2}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-6\right)\left(x-2\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow x^2-8x+12=32\)

\(\Leftrightarrow x^2-8x-20=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)

\(\Leftrightarrow x=-2\) or x=10(đều thỏa)

Vậy ...

ĐKXĐ: x khác 2;3;4;5;6

\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-2}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{x+6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow32=x^2-8x+12\)

\(\Leftrightarrow x^2+8x-20=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)

a: \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

=>(x+4)(x+7)=54

=>x^2+11x+28-54=0

=>(x+13)(x-2)=0

=>x=-13 hoặc x=2

b: \(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-...+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{3}\)

=>\(\dfrac{x+5-x-1}{\left(x+5\right)\left(x+1\right)}=\dfrac{1}{3}\)

=>x^2+6x+5=12

=>x^2+6x-7=0

=>(x+7)(x-1)=0

=>x=-7 hoặc x=1

\(\Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{x+6-x-3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow x^2+9x+18=54\)

\(\Leftrightarrow x^2+9x-36=0\)

=>(x+12)(x-3)=0

=>x=-12 hoặc x=3

\(ĐKXĐ:x\ne-3,-4,-5,-6\)

\(\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{x+6-x-3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\\ \Leftrightarrow\dfrac{3}{x^2+9x+18}=\dfrac{1}{18}\\ \Leftrightarrow x^2+9x+18=54\)

\(\Leftrightarrow x^2+9x-36=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-12\left(tm\right)\end{matrix}\right.\)

Hướng dẫn:

ĐKXĐ:...

Ta có:

\(\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{\left(x+4\right)-\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}+\dfrac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\dfrac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+3}-\dfrac{1}{x+6}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+3\right)\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\left(x+3\right)\left(x+6\right)=54\Leftrightarrow...\)

Sửa lại đề nha:

 \(\dfrac{1}{x^2+9x+12}thành\dfrac{1}{x^2+9x+20}\)

⇔ \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

⇔ \(\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

⇔ \(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)

⇔ \(\dfrac{4}{x^2+8x+12}=\dfrac{1}{8}\)

⇔ \(x^2+8x+12=32\)

⇔ \(x^2+8x-20=0\)

⇔ \(\left(x-2\right)\left(x+10\right)=0\)

⇔ \(\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

b) \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+6\right)+7\left(x+6\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{x+7}{\left(x+4\right)\left(x+7\right)}-\dfrac{x+4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)

\(\Leftrightarrow x^2+11x+28-54=0\)

\(\Leftrightarrow x^2-2x+13x-26=0\)

\(\Leftrightarrow x\left(x-2\right)+13\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\) x - 2 = 0 hoặc x + 13 = 0

\(\Leftrightarrow\) x = 2 hoặc x = -13

Vậy x = 2 hoặc x = -13.