Thu gọn biểu thức:

d) \(\left( {x + 3y} \right)\left( {x - 2y} \right) - \left( {{x^4}y - 6{x^2}{y^3}} \right):{x^2}y\).

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 \[x = \frac{1}{2}\].

a) \(3{x^2} - \sqrt 3 x + \frac{1}{4}\)

\( = {\sqrt 3 ^2}.{x^2} - 2.\sqrt 3 x.\frac{1}{2} + {\left( {\frac{1}{2}} \right)^2}\)

\( = {\left( {\sqrt 3 x} \right)^2} - 2.\sqrt 3 x.\frac{1}{2} + {\left( {\frac{1}{2}} \right)^2}\)

\( = {\left( {\sqrt 3 x - \frac{1}{2}} \right)^2}\).