Đáp án:
Giải thích các bước giải:
$a,xy^2.\sqrt{\dfrac{3}{x^2y^4}}$
$x^2y^4\geq0∀x;y⇒\dfrac{3}{x^2y^4}\geq0∀x;y$
$xy^2.\sqrt{\dfrac{3}{x^2y^4}}$
$=xy^2.\left |\dfrac{\sqrt{3}}{xy^2} \right |$
$=xy^2.\left |\dfrac{\sqrt{3}}{xy^2} \right |(x>0;y\neq0)$
$=\sqrt{3}$
$b,xy^2.\sqrt{\dfrac{3}{x^2y^4}}$
$x^2y^4\geq0∀x;y⇒\dfrac{3}{x^2y^4}\geq0∀x;y$
$xy^2.\sqrt{\dfrac{3}{x^2y^4}}$
$=xy^2.\left |\dfrac{\sqrt{3}}{xy^2} \right |$
$=xy^2.\left |\dfrac{\sqrt{3}}{xy^2} \right | (x<0;y\neq0)$
$=-\sqrt{3}$
$c,\sqrt{\dfrac{27(x-4)^2}{48}}$
$=\sqrt{\dfrac{3.9(x-4)^2}{3.16}}$
$=\sqrt{\dfrac{9(x-4)^2}{16}}$
$=\left |\dfrac{3(x-4)}{4} \right |$
$=\dfrac{3x-12}{4} (x>4)$
$d,\sqrt{\dfrac{27(x-4)^2}{48}}$
$=\sqrt{\dfrac{3.9(x-4)^2}{3.16}}$
$=\sqrt{\dfrac{9(x-4)^2}{16}}$
$=\left |\dfrac{3(x-4)}{4} \right |$
$=\dfrac{-3x+12}{4} (x<4)$
