Question

Hello.

Answer 1

Answer:

$\large{\boxed{\sf 2y^2\sqrt{2y}}}$

Step-by-step explanation:

Here the given expression to us is ;

$\sf\qquad\longrightarrow \dfrac{\sqrt{64xy^5}}{\sqrt{8x}}$

Recall that ,

$\sf\qquad\longrightarrow \dfrac{\sqrt{x}}{\sqrt{y}}=\sqrt{\dfrac{x}{y}}$

On using this , we have ;

$\sf\qquad\longrightarrow \sqrt{\dfrac{\cancel{64}\cancel{x}y^5}{\cancel{8x}}}$

Simplify ,

$\sf\qquad\longrightarrow \sqrt{ 8 y^5}$

The prime factorisation of 8 is 2³ . So ;

$\sf\qquad\longrightarrow \sqrt{ (2^3)(y^5)} =\sqrt{(2^2)(2)(y^2)(y^2)(y)}$

Simplify the square root ,

$\sf\qquad\longrightarrow \pink{ 2y^2\sqrt{2y}}$

Hence the required answer is 2y²√{2y} .

Answer 2

$\\ \rm\rightarrowtail \dfrac{\sqrt{64xy^5}}{\sqrt{8x}}$

$\\ \rm\rightarrowtail \dfrac{\sqrt{8^2xy^5}}{\sqrt{8x}}$

Follow the rule

$\boxed{\Large{\sf \dfrac{a^m}{a^n}=a^{m-n}}}$

$\\ \rm\rightarrowtail \sqrt{8^{2-1}x^{1-1}y^5}$

$\\ \rm\rightarrowtail \sqrt{8^1x^0y^5}$

• a^0=1
• a^1=a

$\\ \rm\rightarrowtail \sqrt{8y^5}$

$\\ \rm\rightarrowtail \sqrt{2^2(2)y^2y^3}$

$\\ \rm\rightarrowtail \sqrt{2^2y(y)(y)(y)}$

$\\ \rm\rightarrowtail 2y^2\sqrt{2y}$