a)
$(2x+1)^2 . 2 = 32$
⇒ $(2x+1)^2 = 16$
⇒ $(2x+1)^2 = 4^2$
⇒ \(\left[ \begin{array}{l}2x+1 = 4\\2x+1 = (-4)\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x = \dfrac{3}{2}\\x = \dfrac{-5}{2}\end{array} \right.\)
Vậy `x ∈ { \frac{3}{2} ; \frac{-5}{2}}`
b)
`|\frac{3}{2}x + \frac{1}{2} | - \frac{3}{5} = \frac{2}{5}`
⇒ `|\frac{3}{2}x + \frac{1}{2} | = 1`
⇒ \(\left[ \begin{array}{l}\dfrac{3}{2}x+ \dfrac{1}{2} =1\\\dfrac{3}{2}x+ \dfrac{1}{2} =-1\end{array} \right.\) ⇒ \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-1\end{array} \right.\)
Vậy `x ∈ { frac{1}{3} ; (-1)}`
#Winner112
