Mn giúp em với ạ em cám ơn mn nhiều

 `b) | x + 1/2 | - 6 = - 2`

`=> | x + 1/2 | = - 2 + 6`

`=> | x + 1/2 | = 4`

`=>` \(\left[ \begin{array}{l}x + \frac{1}{2} = 4\\x + \frac{1}{2} = - 4\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x = \frac{7}{2}\\x = \frac{- 9}{2}\end{array} \right.\) 

Vậy `, x in { 7/2 ; ( - 9 )/2 } .`

`c) ( \sqrt{9} + \sqrt{4} ) . \sqrt{x} = 10 ( ĐK : x ≥ 0 )`

`=> ( 3 + 2 ) . \sqrt{x} = 10`

`=> 5 . \sqrt{x} = 10`

`=> \sqrt{x} = 10 : 5`

`=> \sqrt{x} = 2`

`=> x = 2^2`

`=> x = 4`

Vậy `, x = 4 .`

`d) ( 2x - 3 )^2 = 9`

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

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

`=>` \(\left[ \begin{array}{l}2x = 6\\2x = 0\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x = 3\\x = 0\end{array} \right.\) 

Vậy `, x in { 3 ; 0 } .`