Tìm x, biết: (a)(x−3)^2−x2+4x=5 (b) 4x^2+x−(2x−1)(2x+1)=−5

a) \({\left( {x - 3} \right)^2} - {x^2} + 4x = 5\)

\({x^2} - 6x + 9 - {x^2} + 4x = 5\)

\( - 2x + 9 = 5\)

\( - 2x = - 4\)

\(x = 2\)

Vậy \(x = 2\).

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

\(\begin{array}{l}4{x^2} + x - \left( {4{x^2} - 1} \right) = - 5\\4{x^2} + x - 4{x^2} + 1 = - 5\\x + 1 = - 5\\x = - 6\end{array}\)

Vậy x = \( - \)6

c) \[{\left( {x + 1} \right)^2} + 8\left( {x + 2} \right) = 28\]

\[{x^2} + 2x + 1 + 8x + 16 = 28\]

\[{x^2} + 10x = 11\]

\[{x^2} + 10x + 25 = 36\]

\[{\left( {x + 5} \right)^2} = 36\]

\[x + 5 = 6\] hoặc \[x + 5 = - 6\]

\[x = 1\] hoặc \[x = - 11\]

Vậy \(x \in \left\{ {1; - 11} \right\}\)