a.
`((1)/(9))/(2x+3)=(2x+3)/((25)/(4))`
`→(1)/(9).(25)/(4)=(2x+3)^2`
`→(25)/(36)=(2x+3)^2`
`→(2x+3)^2=((5)/(6))^2`
`→`\(\left[ \begin{array}{l}2x+3=\dfrac{5}{6}\\2x+3=\dfrac{-5}{6}\end{array} \right.\)
`→`\(\left[ \begin{array}{l}2x=\dfrac{-13}{6}\\2x=\dfrac{-23}{6}\end{array} \right.\)
`→`\(\left[ \begin{array}{l}x=\dfrac{-13}{12}\\x=\dfrac{-23}{12}\end{array} \right.\)
Vậy `x=(-13)/(12)` hoặc `x=(-23)/(12)`
b.
`2(2x-3)-2(x+2)-x-2=3(x-5)`
`→4.x-6-2x-4-x-2=3x-15`
`→4.x-6-2x-4-x-2-3x+15=0`
`→(4x-2x-x-3x)+(-6-4-2+15)=0`
`→-2x+3=0`
`→-2x=-3`
`→x=(-3)/(-2)`
`→x=(3)/(2)`
Vậy `x=(3)/(2)`
