ĐKXĐ : Tự tìm nha : )

Ta có : \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)

=> \(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

=> \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)

=> \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

=> \(\frac{x+8}{\left(x+1\right)\left(x+8\right)}-\frac{x+1}{\left(x+8\right)\left(x+1\right)}=\frac{1}{14}\)

=> \(14\left(x+8-x-1\right)=\left(x+1\right)\left(x+8\right)\)

=> \(x^2+x+8x+8=98\)

=> \(x^2+9x-90=0\)

=> \(\left(x+15\right)\left(x-6\right)=0\)

=> \(\left[{}\begin{matrix}x+15=0\\x-6=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-15\\x=6\end{matrix}\right.\) ( TM )

Vậy phương trình trên có nghiệm là \(S=\left\{6,-15\right\}\)

Ta có : \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\) \(\frac{1}{x^2+15x+56}=\frac{1}{14}\)

<=>\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)+...+ \(\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}\)= \(\frac{1}{14}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

<=> \(\frac{x+8-x-1}{\left(x+1\right)\left(x+8\right)}=\frac{1}{14}\)

<=>\(\frac{7.14}{14\left(x+1\right)\left(x+8\right)}=\frac{\left(x+1\right)\left(x+8\right)}{14\left(x+1\right)\left(x+8\right)}\)

<=> \(x^2+9x+8=98\)<=> \(x^2+9x-90=0\)

<=> (x-6)(x+15) =0 

<=> \(\orbr{\begin{cases}x=6\\x=-15\end{cases}}\)

Vậy phương trình có 2 nghiệm  x  \(\in\left(6,15\right)\)

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- Đề ko rõ ràng , lần sau nhớ ghi yêu cầu ?  

Lời giải:

PT \(\Leftrightarrow \frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+....+\frac{1}{(x+7)(x+8)}=\frac{1}{14}\)

(ĐK: $x\neq -1;-2;...;-8$)

\(\Leftrightarrow \frac{(x+2)-(x+1)}{(x+1)(x+2)}+\frac{(x+3)-(x+2)}{(x+2)(x+3)}+....+\frac{(x+8)-(x+7)}{(x+7)(x+8)}=\frac{1}{14}\)

\(\Leftrightarrow \frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)

\(\Leftrightarrow \frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\Leftrightarrow \frac{7}{x^2+9x+8}=\frac{1}{14}\)

\(\Rightarrow x^2+9x+8=98\Leftrightarrow x^2+9x-90=0\Rightarrow x=6\) hoặc $x=-15$ (đều thỏa mãn)

Vậy........

\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

Làm nốt

2/ 

\(T=8x^2-4x+\frac{1}{4x^2}+15\)

\(=\left(4x^2-4x+1\right)+\left(4x^2+\frac{1}{4x^2}-2\right)+16\)

\(=\left(2x-1\right)^2+\left(\frac{4x^2-1}{2x}\right)^2+16\ge16\)

Để PT đc xác định : \(x^2+3x+2\ne0;x^2+5x+6\ne0;.....;x^2+15x+56\ne0\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\ne0;\left(x+2\right)\left(x+3\right)\ne0;....;\left(x+7\right)\left(x+8\right)\ne0\)

\(\Rightarrow x+1;x+2;x+3;....;x+8\ne0\)

\(\Rightarrow x\ne\left\{-8;-7;...;-3;-2;-1\right\}\)

TXĐ : \(x\ne\left\{-8;-7;...;-3;-2;-1\right\}\)

\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+....+\frac{1}{x^2+15x+56}=\frac{1}{14}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+....+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

\(\Leftrightarrow\frac{7}{x^2+9x+8}=\frac{1}{14}\)

\(\Leftrightarrow x^2+9x+8=98\)

\(\Leftrightarrow x^2+9x-90=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+15\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=6\\x=-15\end{cases}}\)(TMĐKXĐ)

Vậy \(x=6\) hoặc \(x=-15\)

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

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

\(\Leftrightarrow\frac{1}{x-1}-\frac{1}{x-5}=\frac{1}{8}\)

\(\Leftrightarrow\frac{x-5-x+1}{\left(x-1\right)\left(x-5\right)}=\frac{1}{8}\)

\(\Leftrightarrow-4.8=x^2-6x+5\)

\(\Leftrightarrow x^2-6x+37=0\)

bo tay

kb với tôi đi

         \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)

\(\Leftrightarrow\frac{\frac{5x+6-2x}{5}}{14}-\frac{x+4}{24}=\frac{\frac{35x+10+9-3x}{5}}{12}+\frac{2}{3}\)

\(\Leftrightarrow\frac{\frac{3x+6}{5}}{14}-\frac{x+4}{24}=\frac{\frac{32x+19}{5}}{12}+\frac{2}{3}\)

\(\Leftrightarrow\left(\frac{3x+6}{5}\cdot\frac{1}{14}\right)-\frac{x+4}{24}=\left(\frac{32x+19}{5}\cdot\frac{1}{12}\right)+\frac{2}{3}\)(CHIA CHO 14 LÀ NHÂN NGHỊCH ĐẢO VỚI 1/14,)                                                                                                                           (CHIA CHO 12 LÀ NHÂN NGHỊCH ĐẢO VỚI 1/12)\(\Leftrightarrow\frac{3x+6}{70}-\frac{x+4}{24}-\frac{32x+19}{60}-\frac{2}{3}=0\)\(\Leftrightarrow\frac{12\left(3x+6\right)-35\left(x+4\right)-14\left(32x+19\right)-2\cdot280}{840}=0\)

 \(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)-14\left(32x+19\right)-560=0\)

\(\Leftrightarrow36x+72-35x-140-448x-266-560=0\)

 \(\Leftrightarrow-447x-894=0\Leftrightarrow x=\frac{-894}{447}=-2\)(NHẬN)

 Vậy tập nghiệm của phương trình là : S = { -2 }

tk cho mk nka ! ! ! th@nks ! ! !