`a) -2/3 : x + 5/8 = - 7/12`

`⇔ -2/3 : x = - 7/12 - 5/8`

`⇔ - 2/3 : x = - 29/24`

`⇔ x = - 2/3 : ( - 29/2 )`

`⇔ x = 16/29`

Vậy `, x = 16/29 .`

`b) ( 2x + 3 )^2 = 25`

`⇔` \(\left[ \begin{array}{l}( 2x + 3 )^2 = 5^2\\( 2x + 3 )^2 = ( - 5 )^2\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}2x + 3 = 5\\2x + 3 = - 5\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}2x = 2\\2x = - 8\end{array} \right.\)  

`⇔` \(\left[ \begin{array}{l}x = 1\\x = - 4\end{array} \right.\) 

Vậy `, x ∈ { 1 ; - 4 }`

bài 1:

`a) -2/3 : x + 5/8=-7/12`

`⇔ -2/3 : x = -19/24`

`⇔ x=16/29` 

Vậy `x=16/29`

`b) (2x+3)^2=25`

`⇔ (2x+3)^2=5^2=(-5)^2`

⇔ \(\left[ \begin{array}{l}2x+3=5\\2x+3=-5\end{array} \right.\)

⇔ \(\left[ \begin{array}{l}2x=2\\2x=-8\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=1\\x=-4\end{array} \right.\) 

Vậy `x∈{1;-4}`