Tìm x biết: (5x-1) ^2 - ( 5x-4 ) ( 5x+4 ) =7

Ta có \({\left( {5x - 1} \right)^2} - \left( {5x - 4} \right)\left( {5x + 4} \right) = 7\)

\(25{x^2} - 10x + 1 - \left( {25{x^2} - 16} \right) = 7\)

\(25{x^2} - 10x + 1 - 25{x^2} + 16 = 7\)

$-10x=-10$

\(x = 1\)

Vậy \(x = 1\).