Cho \(M = 2x - 3 - \sqrt {4{x^2} - 12x + 9} \).

a) Rút gọn M.

b) Tính M khi \(x = \frac{5}{2};\,x = \frac{{ - 1}}{5}\).

Lời giải

a) \(M = 2x - 3 - \sqrt {4{x^2} - 12x + 9} = 2x - 3 - \sqrt {{{\left( {2x - 3} \right)}^2}} \)

\( = 2x - 3 - \left| {2x - 3} \right| = \left\{ \begin{array}{l}2x - 3 - 2x + 3,\,\,\,\,khi\,\,x \ge \frac{3}{2}\\2x - 3 + 2x - 3,\,\,\,\,khi\,x < \frac{3}{2}\end{array} \right.\)

\( = \left\{ \begin{array}{l}0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,x \ge \frac{3}{2}\\4x - 6,\,\,\,\,khi\,x < \frac{3}{2}\end{array} \right.\)

b) Với \(x = \frac{5}{2}\), ta có M = 0.

Với \(x = \frac{{ - 1}}{5}\), ta có \(M = 4.\left( {\frac{{ - 1}}{5}} \right) - 6 = - \frac{{34}}{5}\).

Vậy M = 0 khi \(x = \frac{5}{2}\) và \(M = - \frac{{34}}{5}\) khi \(x = \frac{{ - 1}}{5}\).