\(\dfrac{8}{x}-8+\dfrac{11}{x}-11=\dfrac{9}{x}-9+\dfrac{10}{x}-10\)\(\Leftrightarrow\dfrac{8}{x}+\dfrac{11}{x}-\dfrac{9}{x}-\dfrac{10}{x}=8+11-9-10\)

\(\Leftrightarrow\dfrac{8+11-9-10}{x}=0\)

\(\Leftrightarrow\dfrac{0}{x}=0\)

\(\Leftrightarrow x=0\)

S=\(\left\{0\right\}\)

\(\frac{x+5}{13}+\frac{x+6}{12}+\frac{x+7}{11}=\frac{x+8}{10}+\frac{x+9}{9}+\frac{x+10}{8}\)

\(\Leftrightarrow\left(\frac{x+5}{13}+1\right)+\left(\frac{x+6}{12}+1\right)+\left(\frac{x+7}{11}+1\right)=\left(\frac{x+8}{10}+1\right)+\left(\frac{x+9}{9}+1\right)+\left(\frac{x+10}{8}\right)\)

\(\Leftrightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}=\frac{x+18}{10}+\frac{x+18}{9}+\frac{x+18}{8}\)

ta chuyển về vế trái được 

\(\Leftrightarrow\left(x+18\right)\left(\frac{1}{13}+\frac{1}{122}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)

\(\Leftrightarrow x+2018=0\)(do cái còn lại khác 0)

\(\Leftrightarrow x=-2018\)

mình nghĩ đề cậu viết thiếu mình sửa rồi

Ta có:

\(\frac{x+5}{13}+\frac{x+6}{12}+\frac{x+7}{11}=\frac{x+8}{10}+\frac{x+9}{9}+\frac{x+10}{8}\)

\(\Rightarrow\left(\frac{x+5}{13}+1\right)+\left(\frac{x+6}{12}+1\right)+\left(\frac{x+7}{11}+1\right)=\left(\frac{x+8}{10}+1\right)+\left(\frac{x+9}{9}+1\right)+\left(\frac{x+10}{8}+1\right)\)

\(\Rightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}=\frac{x+18}{10}+\frac{x+18}{9}+\frac{x+18}{8}\)

\(\Rightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}-\frac{x+18}{10}-\frac{x+18}{9}-\frac{x+18}{8}=0\)

\(\Rightarrow\left(x+18\right)\times\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)

Vì \(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)

\(\Rightarrow x+18=0\)

\(\Rightarrow x=-18\)

Vậy phương trình có nghiệm là x = -18

xin lỗi vì tớ nhầm một chút ở dòng cuối cho mình xin lỗi nhé

x = -15

1 đúng nhá

x = -15

1 đúng nhá

\(\dfrac{x-187}{13}+\dfrac{x-170}{15}+\dfrac{x-149}{17}+\dfrac{x-124}{19}=10\)

`<=>(x-187)/13+(x-170)/15+(x-149)/17+(x-124)/19-10=0`

`<=>(x-187)/13-1+(x-170)/15-2+(x-149)/17-3+(x-124)/19-4=0`

`<=>(x-200)/13+(x-200)/15+(x-200)/17+(x-200)/19=0`

`<=>(x-200)(1/13+1/15+1/17+1/19)=0`

`<=>x-200=0(1/13+1/15+1/17+1/19>0)`

`<=>x=200`

\(=>\left(\dfrac{x-187}{13}-1\right)+\left(\dfrac{x-170}{15}-2\right)+\left(\dfrac{x-149}{17}-3\right)+\left(\dfrac{x-124}{19}-4\right)=0\)\(< =>\left(\dfrac{x-187}{13}-\dfrac{13}{13}\right)+\left(\dfrac{x-170}{15}-\dfrac{30}{15}\right)+\left(\dfrac{x-149}{17}-\dfrac{51}{17}\right)+\left(\dfrac{x-124}{19}-\dfrac{76}{19}\right)=0\)

\(< =>\left(\dfrac{x-200}{13}\right)+\left(\dfrac{x-200}{15}\right)+\left(\dfrac{x-200}{17}\right)+\left(\dfrac{x-200}{19}\right)=0\)

\(< =>\left(x-200\right)\left(\dfrac{1}{13}+\dfrac{1}{15}+\dfrac{1}{17}+\dfrac{1}{19}\right)=0\)

\(< =>x-200=0\)

<=>x=200

=>\(\left(\dfrac{x-187}{13}-1\right)+\left(\dfrac{x-170}{15}-2\right)+\left(\dfrac{x-149}{17}-3\right)+\left(\dfrac{x-124}{19}-4\right)=0\)

=>x-200=0

=>x=200

     \(\frac{149-x}{25}+\frac{170-x}{23}+\frac{187-x}{21}+\frac{200-x}{19}=10\)

\(\Rightarrow\frac{149-x}{25}-1+\frac{170-x}{23}-2+\frac{187-x}{21}-3+\frac{200-x}{19}-4=0\)

\(\Rightarrow\frac{124-x}{25}+\frac{124-x}{23}+\frac{124-x}{21}+\frac{124-x}{19}=0\)

\(\Rightarrow\left(124-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)

Mà \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}>0\Rightarrow x-124=0\Rightarrow x=124\)

\(\dfrac{x^2-26}{10}+\dfrac{x^2-25}{11}\ge\dfrac{x^2-24}{12}+\dfrac{x^2-23}{13}\)

\(\Leftrightarrow\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)

\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)

\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)

\(\Leftrightarrow\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)

Vì \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\Rightarrow x^2-36\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-6\\x\ge6\end{matrix}\right.\)

Bất phương trình đó tương đương với:

 \(\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)

⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)

⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)

⇔ \(\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)

+)Vì \(\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}\) nên \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\) 

⇔ \(x^2-36\ge0\)

⇔ \(x^2\ge36\)

⇔ \(\sqrt{x^2}\ge6\)

⇔ \(\left|x\right|\ge6\)

⇔ \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)

➤ Vậy \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)

\(\frac{x-7}{x-8}-\frac{x-8}{x-9}=\frac{x-10}{x-11}-\frac{x-11}{x-12}\)

\(\frac{x-7}{x-8}-\frac{x-8}{x-9}-\frac{x-10}{x-11}+\frac{x-11}{x-12}=0\)

Rồi còn lại làm típ