$a,\\\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\\\leftrightarrow \dfrac{x-23}{24}+\dfrac{x-23}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0\\\leftrightarrow(x-23)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\\\text{Nhận thấy:}\\\dfrac{1}{24}>\dfrac{1}{26}\to \dfrac{1}{24}-\dfrac{1}{26}>0\\\dfrac{1}{25}>\dfrac{1}{27}\to \dfrac{1}{25}-\dfrac{1}{27}>0\\\to \dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}>0\\\to x-23=0\\\leftrightarrow x=23$
$b,\\\left(\dfrac{x+2}{98}+1 \right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\\\leftrightarrow \dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\\\leftrightarrow \dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+1000}{95}=0\\\leftrightarrow (x+100)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\\\text{Nhận thấy:}\\\dfrac{1}{98}<\dfrac{1}{96}\to \dfrac{1}{98}-\dfrac{1}{96}<0\\\dfrac{1}{97}<\dfrac{1}{95}\to \dfrac{1}{97}-\dfrac{1}{95}<0\\\to \dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}<0\\\to x+100=0\\\leftrightarrow x=-100$
$c,\\\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\\\leftrightarrow \left(\dfrac{x+1}{2004}+1\right)+\left(\dfrac{x+2}{2003}+1\right)=\left(\dfrac{x+3}{2002}+1\right)+\left(\dfrac{x+4}{2001}+1\right)\\\leftrightarrow \dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\\\leftrightarrow \dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}-\dfrac{x+2005}{2002}-\dfrac{x+2005}{2001}=0\\\leftrightarrow (x+2005)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\\\text{Nhận thấy:}\\\dfrac{1}{2004}<\dfrac{1}{2002}\to \dfrac{1}{2004}-\dfrac{1}{2002}<0\\\dfrac{1}{2003}<\dfrac{1}{2001}\to \dfrac{1}{2003}-\dfrac{1}{2001}<0\\\to \dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}<0\\\to x+2005=0\\\leftrightarrow x=-2005$
$d,\\\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\\\leftrightarrow \left(\dfrac{201-x}{99}+1\right)+\left(\dfrac{203-x}{97}+1\right)+\left(\dfrac{205-x}{95}+1\right)=0\\\leftrightarrow \dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\\\leftrightarrow (300-x)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\\\text{Vì} \ \dfrac{1}{99}+\dfrac1{97}+\dfrac1{95}>0\\\to 300-x=0\\\leftrightarrow x=300$
$e,\\\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\\\leftrightarrow \left(\dfrac{x-45}{55}-1\right)+\left(\dfrac{x-47}{53}-1\right)=\left(\dfrac{x-55}{45}-1\right)+\left(\dfrac{x-53}{47}-1\right)\\\leftrightarrow \dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\\\leftrightarrow \dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\\\leftrightarrow (x-100)\left(\dfrac1{55}+\dfrac1{53}-\dfrac1{45}-\dfrac1{47}\right)=0\\\text{Nhận thấy:}\\\dfrac1{55}<\dfrac1{47}\to \dfrac{1}{55}-\dfrac{1}{47}<0\\\dfrac{1}{53}<\dfrac{1}{45}\to \dfrac{1}{53}-\dfrac{1}{45}<0\\\to \dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}<0\\\to x-100=0\\\leftrightarrow x=100 $
$f,\\\dfrac{x+1}9+\dfrac{x+2}{8}=\dfrac{x+3}7+\dfrac{x+4}6\\\leftrightarrow \left(\dfrac{x+1}9+1\right)+\left(\dfrac{x+2}8+1\right)=\left(\dfrac{x+3}{7}+1\right)+\left(\dfrac{x+4}6+1\right)\\\leftrightarrow \dfrac{x+10}9+\dfrac{x+10}8=\dfrac{x+10}7+\dfrac{x+10}6\\\leftrightarrow \dfrac{x+10}{9}+\dfrac{x+10}{8}-\dfrac{x+10}{7}-\dfrac{x+10}{6}=0\\\leftrightarrow (x+10)\left(\dfrac19+\dfrac18-\dfrac17-\dfrac16\right)=0\\\text{Nhận thấy:}\\\dfrac19<\dfrac17\to \dfrac19-\dfrac17<0\\\dfrac18<\dfrac16\to \dfrac18-\dfrac16<0\\\dfrac19+\dfrac18-\dfrac17-\dfrac16<0\\\to x+10=0\\\leftrightarrow x=-10$
$g,\\\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\\\leftrightarrow \left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+4}{96}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+8}{92}+1\right)\\\leftrightarrow \dfrac{x+100}{98}+\dfrac{x+100}{96}=\dfrac{x+100}{94}+\dfrac{x+100}{92}\\\leftrightarrow \dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\\\leftrightarrow (x+100)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\\\text{Nhận thấy:}\\\dfrac1{98}<\dfrac1{94}\to \dfrac{1}{98}-\dfrac{1}{94}<0\\\dfrac{1}{96}<\dfrac{1}{92}\to \dfrac{1}{96}-\dfrac{1}{92}<0\\\to \dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}<0\\\to x+100=0\\\leftrightarrow x=-100$
$h,\\\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\\\leftrightarrow \dfrac{2-x}{2002}+1=\left(\dfrac{1-x}{2003}+1\right)-\left(\dfrac{x}{2004}-1\right)\\\leftrightarrow \dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}\\\leftrightarrow \dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}=0\\\leftrightarrow (2004-x)\left(\dfrac{1}{2002}-\dfrac{1}{2003}+\dfrac{1}{2004}\right)=0\\\text{Nhận thấy:}\\\dfrac{1}{2002}>\dfrac{1}{2003}\to \dfrac{1}{2002}-\dfrac1{2003}>0\\\to \dfrac{1}{2002}-\dfrac{1}{2003}+\dfrac1{2004}>0\\\to 2004-x=0\\\leftrightarrow x=2004$
$i,\\\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\\\leftrightarrow \left(\dfrac{x^2-10x-29}{1971}-1\right)+\left(\dfrac{x^2-10x-27}{1973}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\\\leftrightarrow \dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\\\leftrightarrow \dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-2000}{29}-\dfrac{x^2-10x-2000}{27}=0\\\leftrightarrow (x^2-10x-2000)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\\\text{Nhận thấy:}\\\dfrac{1}{1971}<\dfrac{1}{29}\to \dfrac{1}{1971}-\dfrac{1}{29}<0\\\dfrac{1}{1973}<\dfrac{1}{27}\to \dfrac{1}{1973}-\dfrac{1}{27}<0\\\to \dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}<0\\\to x^2-10x-2000=0\\\leftrightarrow x^2-10x+25-2025=0\\\leftrightarrow x^2-10x+25=2025\\\leftrightarrow (x-5)^2=45^2\\\leftrightarrow \left[\begin{array}{l}x-5=-45\\x-5=45\end{array}\right. \leftrightarrow \left[\begin{array}{l}x=-40\\x=50\end{array}\right.$
