Lời giải

Đáp án đúng là: C

$A = \left( {4{x^2} - 16} \right).\frac{{7x - 2}}{{3x + 6}} = \frac{{\left( {4{x^2} - 16} \right)\left( {7x - 2} \right)}}{{3x + 6}}$

$= \frac{{4\left( {x - 2} \right)\left( {x + 2} \right)7x - 2}}{{3\left( {x + 2} \right)}}$$= \frac{{4\left( {x - 2} \right)7x - 2}}{3}$

$= \frac{4}{3}\left( {7{x^2} - 2x - 14x + 4} \right) = \frac{4}{3}\left( {7{x^2} - 16x + 4} \right)$$= \frac{4}{3}\left[ {{{\left( {\sqrt 7 x} \right)}^2} - 2.\sqrt 7 x.\frac{8}{{\sqrt 7 }} + {{\left( {\frac{8}{{\sqrt 7 }}} \right)}^2} + 4 - {{\left( {\frac{8}{{\sqrt 7 }}} \right)}^2}} \right]$

$= \frac{4}{3}\left[ {{{\left( {\sqrt 7 x - \frac{8}{{\sqrt 7 }}} \right)}^2} - \frac{{36}}{7}} \right]$.

Ta có: ${\left( {\sqrt 7 x - \frac{8}{{\sqrt 7 }}} \right)^2} \ge 0\forall x \Rightarrow {\left( {\sqrt 7 x - \frac{8}{{\sqrt 7 }}} \right)^2} - \frac{{36}}{7} \ge - \frac{{36}}{7}\forall x$

$\frac{4}{3}\left[ {{{\left( {\sqrt 7 x - \frac{8}{{\sqrt 7 }}} \right)}^2} - \frac{{36}}{7}} \right] \ge \frac{4}{3}.\left( { - \frac{{36}}{7}} \right) = - \frac{{48}}{7}$ hay $A \ge - \frac{{48}}{7}$

Dấu “=” xảy ra khi và chỉ khi ${\left( {\sqrt 7 x - \frac{8}{{\sqrt 7 }}} \right)^2} = 0$ hay $x = \frac{8}{7}$.