`h)(4x+3)^2=49`

`<=>(4x+3)^2=7^2`

`<=>`\(\left[ \begin{array}{l}4x+3=7\\4x+3=-7\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=1\\x=\frac{-5}{2}\end{array} \right.\)

Vậy `S={1;(-5)/2}`

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`i)(3x+5)^2-(x-2)^2=0`

`<=>(3x+5-x+2)(3x+5+x-2)=0`

`<=>(2x+7)(4x+3)=0`

`<=>`\(\left[ \begin{array}{l}2x+7=0\\4x+3=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=\frac{-7}{2}\\x=\frac{-3}{4}\end{array} \right.\)

Vậy `S={(-7)/2;(-3)/4}`

`@` Áp dụng HĐT:`a^2-b^2=(a-b)(a+b)`

$(4x+3)^2=49$

$⇔(4x+3)^2=(±7)^2$

$⇔$ \(\left[ \begin{array}{l}4x+3=7\\4x+3=-7\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}4x=4\\4x=-10\end{array} \right.\) 

$⇔$ \(\left[ \begin{array}{l}x=1\\x=-\dfrac{5}{2} \end{array} \right.\) 

Vậy `S={1;-5/2}`

$i,$ $(3x+5)^2-(x-2)^2=0$

$⇔(3x+5-x+2)(3x+5+x-2)=0$

$⇔(2x+7)(4x+3)=0$

$⇔$ \(\left[ \begin{array}{l}x=-\dfrac{7}{2}\\x=-\dfrac{3}{4} \end{array} \right.\) 

Vậy `S={-7/2;-3/4}`