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Sagot :
Answer:
[tex]\large{\boxed{\sf 2y^2\sqrt{2y}}}[/tex]
Step-by-step explanation:
Here the given expression to us is ;
[tex]\sf\qquad\longrightarrow \dfrac{\sqrt{64xy^5}}{\sqrt{8x}}[/tex]
Recall that ,
[tex]\sf\qquad\longrightarrow \dfrac{\sqrt{x}}{\sqrt{y}}=\sqrt{\dfrac{x}{y}}[/tex]
On using this , we have ;
[tex]\sf\qquad\longrightarrow \sqrt{\dfrac{\cancel{64}\cancel{x}y^5}{\cancel{8x}}}[/tex]
Simplify ,
[tex]\sf\qquad\longrightarrow \sqrt{ 8 y^5}[/tex]
The prime factorisation of 8 is 2³ . So ;
[tex]\sf\qquad\longrightarrow \sqrt{ (2^3)(y^5)} =\sqrt{(2^2)(2)(y^2)(y^2)(y)}[/tex]
Simplify the square root ,
[tex]\sf\qquad\longrightarrow \pink{ 2y^2\sqrt{2y}} [/tex]
Hence the required answer is 2y²√{2y} .
[tex]\\ \rm\rightarrowtail \dfrac{\sqrt{64xy^5}}{\sqrt{8x}}[/tex]
[tex]\\ \rm\rightarrowtail \dfrac{\sqrt{8^2xy^5}}{\sqrt{8x}}[/tex]
Follow the rule
[tex]\boxed{\Large{\sf \dfrac{a^m}{a^n}=a^{m-n}}}[/tex]
[tex]\\ \rm\rightarrowtail \sqrt{8^{2-1}x^{1-1}y^5}[/tex]
[tex]\\ \rm\rightarrowtail \sqrt{8^1x^0y^5}[/tex]
- a^0=1
- a^1=a
[tex]\\ \rm\rightarrowtail \sqrt{8y^5}[/tex]
[tex]\\ \rm\rightarrowtail \sqrt{2^2(2)y^2y^3}[/tex]
[tex]\\ \rm\rightarrowtail \sqrt{2^2y(y)(y)(y)}[/tex]
[tex]\\ \rm\rightarrowtail 2y^2\sqrt{2y}[/tex]
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