Tìm \(x,\) biết:

d) \[25{\left( {x + 3} \right)^2} + \left( {1-5x} \right)\left( {1 + 5x} \right) = 8.\]

Lời giải:

a) Ta có \[A = 4\left( {x - 2} \right)\left( {x + 1} \right) + {\left( {2x - 4} \right)^2} + {\left( {x + 1} \right)^2}\]

          \( = {\left( {2x - 4} \right)^2} + 2.2\left( {x - 2} \right)\left( {x + 1} \right) + {\left( {x + 1} \right)^2}\)

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

          \( = {\left[ {\left( {2x - 4} \right) + \left( {x + 1} \right)} \right]^2}\)

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

          \( = {\left( {3x - 3} \right)^2}\)

          \( = {\left[ {3\left( {x - 1} \right)} \right]^2}\)

          \( = 9{\left( {x - 1} \right)^2}\).

Do đó \(A = 9{\left( {x - 1} \right)^2}\).

Thay \[x = \frac{1}{2}\] vào \(A\) ta được \(A = 9{\left( {\frac{1}{2} - 1} \right)^2} = 9.{\left( { - \frac{1}{2}} \right)^2} = 9.\frac{1}{4} = \frac{9}{4}\).

Vậy \(A = \frac{9}{4}\) tại \[x = \frac{1}{2}\].

b) \[{\left( {x + 3} \right)^2} + \left( {5 - x} \right)\left( {5 + x} \right) = - 3\]

\[{x^2} + 6x + 9 + 25 - {x^2} = - 3\]

\[6x + 9 + 25 = - 3\]

\[6x = - 37\]

\(x = \frac{{ - 37}}{6}\)

Vậy \(x = \frac{{ - 37}}{6}\).