a) \(x\sqrt{7}\)
\(\sqrt{7\cdot x^2}\)
\(=\sqrt{7x^2}\)
b) \(x\sqrt{15}\)
\(=-\sqrt{15\cdot x^2}\)
\(=-\sqrt{15x^2}\)
(Do \(x\le0\Rightarrow x\sqrt{15}=-\sqrt{15x^2}\))
c) \(\dfrac{1}{y}\sqrt{19y}\)
\(=\sqrt{\dfrac{1}{y^2}\cdot19y}\)
\(=\sqrt{\dfrac{19y}{y^2}}\)
\(=\sqrt{\dfrac{19}{y}}\)
d) \(\dfrac{1}{3}y\sqrt{\dfrac{27}{y^2}}\)
\(=\dfrac{y}{3}\cdot\sqrt{\dfrac{27}{y^2}}\)
\(=-\sqrt{\dfrac{y^2}{9}\cdot\dfrac{27}{y^2}}\)
\(=-\sqrt{\dfrac{27}{9}}\)
\(=-\sqrt{3}\)
(Do: \(y< 0\Rightarrow\dfrac{y}{3}=-\sqrt{\dfrac{y^2}{9}}\))
a: Vì x>=0 nên \(x\sqrt{7}=\sqrt{x^2\cdot7}\)
b: Vì x<=0 nên \(x\sqrt{15}=-\sqrt{x^2\cdot15}=-\sqrt{15x^2}\)
c: y>0 nên \(\dfrac{1}{y}\sqrt{19y}=\sqrt{\dfrac{19y}{y^2}}=\sqrt{\dfrac{19}{y}}\)
d: Vì y<=0 nên \(\dfrac{1}{3}y\sqrt{\dfrac{27}{y^2}}=-\sqrt{\dfrac{1}{9}y^2\cdot\dfrac{27}{y^2}}=-\sqrt{3}\)
