`a)(2x+7)^2=(x+3)^2`
`→(2x+7)^2-(x+3)^2=0`
`→(2x+7-x-3)(2x+7+x+3)=0`
`→(x+4)(3x+10)=0`
`→` \(\left[ \begin{array}{l}x+4=0\\3x+10=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=-4\\x=\dfrac{-10}{3}\end{array} \right.\)
Vậy `x∈{-4;-10/3}`
`b)(4x+14)^2=(7x+2)^2`
`→(4x+14)^2-(7x+2)^2=0`
`→(4x+14-7x-2)(4x+14+7x+2)=0`
`→(-3x+12)(11x+16)=0`
`→` \(\left[ \begin{array}{l}-3x+12=0\\11x+16=0\end{array} \right.\)
`→` \(\left[ \begin{array}{l}-3x=-12\\11x=-16\end{array} \right.\)
`→` \(\left[ \begin{array}{l}x=4\\x=\dfrac{-16}{11}\end{array} \right.\)
Vậy `x∈{4;-16/11}`
