`a)` `(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0`
`<=>4x^2-4x+1+x^2+6x+9-5(x^2-49)=0`
`<=>5x^2+2x+10-5x^2+245=0`
`<=>2x+255=0`
`<=>2x=-255`
`<=>x=-255/2`
Vậy `x=-255/2`
`b)` `(2x+1)^2-(x+1)^2=0`
`<=>(2x+1+x+1)[2x+1-(x+1)]=0`
`<=>(3x+2)(2x+1-x-1)=0`
`<=>x(3x+2)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\3x+2=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=0\\x=-\dfrac{2}{3}\end{array} \right.\)
Vậy `x=0;x=2/3`