Question

Rút gọn biểu thức 

a)\(\frac{\sqrt{2x^3}}{\sqrt{8x}}\)(x>0)

b)(3−√5)(3+√5)

c)\(\sqrt{\frac{3x^2y^4}{27}}\)(x<0,x=0)

e)\(\frac{y}{x^2}\sqrt{\frac{36x^4}{y^2}}\)(y<0)

f)\(\frac{\sqrt{99999999}}{\sqrt{11111111}}\)

MK CẦN GẤP CÁC BẠN GIÚP MÌNH NHÉ!!! PLEASE

Answer 1

a) \(\frac{\sqrt{2x^3}}{\sqrt{8x}}=\sqrt{\frac{2x^3}{8x}}=\frac{1}{2}x\)

b) \(\left(3-\sqrt{5}\right)\left(x+\sqrt{5}\right)=3^2-\left(\sqrt{5}\right)^2=9-5=4\)

c) \(\sqrt{\frac{3x^2y^4}{27}}=0\)

\(y\ne0\)

Thì \(\sqrt{\frac{3x^2y^4}{27}}=\frac{1}{3}xy^2\)

e) \(\frac{y}{x^2}\sqrt{\frac{36x^4}{y^2}}=\frac{y}{x^2}.\frac{6x^2}{\left|y\right|}=\frac{6y}{\left|y\right|}\)

Vì y < 0 nên \(\left|y\right|=-y\)

Vậy \(\frac{6y}{\left|y\right|}=\frac{6y}{-y}=-6\)

f) \(\frac{\sqrt{99999999}}{\sqrt{11111111}}=\sqrt{\frac{99999999}{11111111}}=\sqrt{9}=3\)