a)\(-ĐKXĐ:\hept{\begin{cases}x-14\ne0;x-13\ne0\\x-9\ne0\\x-11\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne14;x\ne13\\x\ne9\\x\ne11\end{cases}}\)

 - Ta có : \(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\)

\(\Leftrightarrow\frac{2}{x-14}-\frac{5}{x-13}-\frac{2}{x-9}+\frac{5}{x-11}=0\)

\(\Leftrightarrow\left(\frac{2}{x-14}-\frac{2}{x-9}\right)-\left(\frac{5}{x-13}-\frac{5}{x-11}\right)=0\)

\(\Leftrightarrow2\left(\frac{1}{x-14}-\frac{1}{x-9}\right)-5\left(\frac{1}{x-13}-\frac{1}{x-11}\right)=0\)\(\Leftrightarrow2.\frac{\left(x-9\right)-\left(x-14\right)}{\left(x-9\right)\left(x-14\right)}-5.\frac{\left(x-11\right)-\left(x-13\right)}{\left(x-11\right)\left(x-13\right)}=0\)

\(\Leftrightarrow2.\frac{5}{\left(x-9\right)\left(x-14\right)}-5.\frac{2}{\left(x-11\right)\left(x-13\right)}=0\)

\(\Leftrightarrow\frac{10}{\left(x-9\right)\left(x-14\right)}-\frac{10}{\left(x-11\right)\left(x-13\right)}=0\)

\(\Leftrightarrow10\left[\frac{1}{\left(x-9\right)\left(x-14\right)}-\frac{1}{\left(x-11\right)\left(x-13\right)}\right]=0\)

\(\Leftrightarrow\frac{\left(x-11\right)\left(x-13\right)}{\left(x-9\right)\left(x-14\right)\left(x-11\right)\left(x-13\right)}-\frac{\left(x-9\right)\left(x-14\right)}{\left(x-9\right)\left(x-14\right)\left(x-11\right)\left(x-13\right)}=\) \(0\)

\(\Leftrightarrow\left(x-11\right)\left(x-13\right)-\left(x-9\right)\left(x-14\right)=0\)

\(\Leftrightarrow x^2-24x+143-x^2+23x-126=0\)

\(\Leftrightarrow-x+17=0\Leftrightarrow-x=-17\Leftrightarrow x=17\)

Vậy pt có tập nghiệm S = { 17 }

P/s: Mk làm hơi lòng vòng, bn thông cảm nhé !

ở câu a sai đề cho x+14 đúng là x-14 

\(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}.\) \(\left(ĐKXĐ:x\ne14;13;11;9\right)\)

\(\Leftrightarrow\frac{2}{x-14}-\frac{2}{x-9}=\frac{5}{x-13}-\frac{5}{x-11}\)

\(\Leftrightarrow\frac{10}{\left(x-14\right)\left(x-9\right)}=\frac{10}{\left(x-13\right)\left(x-11\right)}\)

\(\Rightarrow\left(x-14\right)\left(x-9\right)=\left(x-13\right)\left(x-11\right)\)

\(\Leftrightarrow x^2-23x+126=x^2-24x+143\)

\(\Leftrightarrow x=17\left(tm\right)\)

Vậy phương trình có tập nhiệm \(S=\left\{17\right\}\)

<=> 10/ [( x - 14)( x - 9 )] = 10 / [(x - 13) ( x - 11)]

<=> ( x - 14) ( x - 9) = ( x - 3) ( x - 11) 

<=> x^2 - 23x + 126 = x^2 - 24x + 143

<=> x = 17 <=> x = 17 ( thỏa được)

Tập nghiệm của PT là {17}