Ta có:
+) 3√5=√32.5=√9.5=√45.35=32.5=9.5=45.
+) −5√2=−√52.2=−√25.2=−√50.−52=−52.2=−25.2=−50.
+) Với xy>0xy>0 thì √xyxy có nghĩa nên ta có:
−23√xy=−√(23)2.xy=−√49xy.−23xy=−(23)2.xy=−49xy.
+) Với x>0x>0 thì √2x2x có nghĩa nên ta có:
x√2x=√x2.2x=√x2.2xx2x=x2.2x=x2.2x=√2x.xx=√2x.
a, \(3\sqrt{5}=\sqrt{9.5}=\sqrt{45}\)
b, \(-5\sqrt{2}=-\sqrt{25.2}=-\sqrt{50}\)
c, \(-\frac{2}{3}\sqrt{xy}=-\sqrt{\frac{4}{9}xy}\)
d, \(x\sqrt{\frac{2}{x}}=\sqrt{\frac{2x^2}{x}}=\sqrt{2x}\)
a) .
b) .
c) Vì do đó biểu thức có nghĩa.
.
d) Với có nghĩa.
\(3\sqrt{5}=\sqrt{9}.\sqrt{5}=\sqrt{45}\)
\(-5\sqrt{2}=-\sqrt{25}.\sqrt{2}=-\sqrt{50}\)
\(-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\dfrac{4}{9}}.\sqrt{xy}=-\sqrt{\dfrac{4xy}{9}}\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2}.\sqrt[]{\dfrac{2}{x}}=\sqrt{2x}\)
\(3\sqrt{5}=\sqrt{9}.\sqrt{5}=\sqrt{45}\)
\(-5\sqrt{2}=-\sqrt{25}.\sqrt{2}=-\sqrt{50}\)
\(\dfrac{-2}{3}\sqrt{xy}=-\sqrt{\dfrac{4}{9}xy}\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2.\dfrac{2}{x}}=\sqrt{2x}\)
\(x\sqrt{\dfrac{2}{x}}=\dfrac{\sqrt{x^2\cdot2}}{x}\)\(3\sqrt{5}=\sqrt{3^2\cdot5}\)
\(-5\sqrt{2}=\sqrt{\left(-5\right)^2\cdot2}\)\(-\dfrac{2}{3}\sqrt{xy}=\sqrt{\left(-\dfrac{2}{3}\right)^2\cdot xy}\)
a)\(\sqrt{45}\)
b)-\(\sqrt{50}\)
c)-\(\sqrt{\dfrac{4xy}{9}}\)
d)\(\sqrt{2x}\)
3\(\sqrt{5}\)=\(\sqrt{45}\)
-5\(\sqrt{2}\)=-\(\sqrt{50}\)
-\(\dfrac{2}{3}\)\(\sqrt{xy}\)=-\(\sqrt{\dfrac{4xy}{9}}\)
\(x\sqrt{\dfrac{2}{x}}\)=\(\sqrt{2x}\)
\(3\sqrt{5}=\sqrt{9\cdot5}=\sqrt{45}\) ;\(-5\sqrt{2}=-\sqrt{25\cdot2}=-\sqrt{50}\)
\(-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\dfrac{4}{9}xy}\left(xy\ge0\right)\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2\cdot\dfrac{2}{x}}=\sqrt{2x}\left(x>0\right)\)
a)3\(\sqrt{5}\)=\(\sqrt{3^2.5}\)=\(\sqrt{9.5}\)=\(\sqrt{45}\)
b)\(-5\sqrt{2=-\sqrt{5^2}.2=\sqrt{25.2=-\sqrt{50}}}\)
c)vì xy \(\ge0\)do đó biểu thức \(\sqrt{xy}\) có nghĩa.
ta có \(-\dfrac{2}{3}\)\(\sqrt{xy=}\)\(-\sqrt{\dfrac{\left(2\right)^2}{\left(3\right)}.xy=}\)\(-\sqrt{\dfrac{4xy}{9}}\)
d) với x>0 thì\(\sqrt{\dfrac{2}{x}}\) có nghĩa.
\(x\sqrt{\dfrac{2}{x}}=\)\(\sqrt{\dfrac{2}{x}x^2}\)=\(\sqrt{2x}\)
a căn 35
b âm căn 50
c âm căn 4xy trên 9
d căn 2x
a)3\(\sqrt{5}\)=\(\sqrt{3^2.5}\)=\(\sqrt{9.5}\)=\(\sqrt{45}\) b)-5\(\sqrt{2}\)= -5\(\sqrt{5^2.2}\)= -\(\sqrt{25.2}\)=-\(\sqrt{50}\) c)vì xy\(\ge0\) do đó biểu thức căn \(\sqrt{xy}\) có nghĩa: Ta có: -\(\dfrac{2}{3}\sqrt{xy}\)=-\(\sqrt{\left(\dfrac{2}{3}\right)^2.xy}\)= -\(\sqrt{\dfrac{4xy}{9}}\) d)với x>0 thì \(\sqrt{\dfrac{2}{x}}\) có nghĩa: x\(\sqrt{\dfrac{2}{x}}\)= \(\sqrt{x^2\dfrac{2}{x}}\) =\(\sqrt{2x}\)
a )\(3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}\) b) \(-5\sqrt{2}=-\sqrt{5^2.2}=-\sqrt{50}\) c ) \(\dfrac{-2}{3}\sqrt{xy}=-\sqrt{\left(\dfrac{2}{3}\right)^2xy}=-\sqrt{\dfrac{4xy}{9}}\) d)\(x\sqrt{\dfrac{2}{x}}=\)
a)\(3\sqrt{5}=\sqrt{9.5}=\sqrt{45}\)
b)\(-5\sqrt{2}=-\sqrt{25.2}=\sqrt{50}\)
c)\(-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\left(\dfrac{2}{3}\right)^2xy}=-\sqrt{\dfrac{4xy}{9}}\)
d)\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2.\dfrac{2}{x}}=\sqrt{2x}\)
a)\(3\sqrt{5}=\sqrt{3^2\cdot5}=\sqrt{9\cdot5}=\sqrt{45}\) b ) \(-5\sqrt{2}=-\sqrt{5^2\cdot2}=-\sqrt{25\cdot2}=-\sqrt{50}\) c)\(xy\ge0\Rightarrow\sqrt{xy}conghia\) \(-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\left(\dfrac{2}{3}\right)^2\cdot xy}=-\sqrt{\dfrac{4\cdot xy}{9}}\) d) \(x>0\Rightarrow\sqrt{\dfrac{2}{x}}conhia\) \(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2\cdot\dfrac{2}{x}}=\sqrt{2x}\)
\(\sqrt{54};-\sqrt{50;}-\sqrt{\dfrac{4}{9}xy};-\sqrt{2x}\)
a)3\(\sqrt{5}\)=\(\sqrt{3^2.5}\)=\(\sqrt{9.5}\)=\(\sqrt{45}\)
b)-5\(\sqrt{2}\)=-\(\sqrt{5^2}.2\)= -\(\sqrt{25.2}\)=-\(\sqrt{50}\)
c)vì xy\(\ge\)0 do đó biểu thức \(\sqrt{xy}\) có nghĩa
Ta có:-\(\dfrac{2}{3}\)\(\sqrt{xy}\)=-\(\sqrt{(\dfrac{2}{3})^2.xy}\)=-\(\sqrt{\dfrac{4xy}{9}}\)
d) với x>0 thì \(\sqrt{\dfrac{2}{x}}\) có nghĩa
x\(\sqrt{\dfrac{2}{x}}\)=\(\sqrt{x^2\dfrac{2}{x}}\)=\(\sqrt{2x}\)
a = \(\sqrt{ }\)3^2.5 = \(\sqrt{ }\)45 b = -\(\sqrt{ }\)5^2.2 = -\(\sqrt{ }\)50 c = -\(\sqrt{ }\)(2/3)^2xy = -\(\sqrt{ }\)4/9xy d = \(\sqrt{ }\)x^2.2/x = \(\sqrt{ }\)2x
a)
b)
c) bởi vì
do đó biểu thức căn xy
Ta có: .
a) .
b) .
c) Vì do đó biểu thức có nghĩa.
Ta có: .
d) Với thì có nghĩa.
.
\(-5\sqrt{2}=-\sqrt{5^2\cdot2}=-\sqrt{50};\)\(3\sqrt{5}=\sqrt{3^2\cdot5}=\sqrt{45};\)\(-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\left(\dfrac{2}{3}\right)^2\cdot xy}=\sqrt{\dfrac{4xy}{9}};\)\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2\cdot\dfrac{2}{x}}=\sqrt{2x};\)
