lke123
lke123 lke123
  • 29-09-2018
  • Mathematics
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Help with math plssssss!

Help with math plssssss class=

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gmany
gmany gmany
  • 29-09-2018

[tex]\dfrac{x\sqrt3}{\sqrt{4x}}=\dfrac{x\sqrt3}{\sqrt4\cdot\sqrt{x}}=\dfrac{x\sqrt3}{2\sqrt{x}}=\dfrac{x\sqrt3\cdot\sqrt{x}}{2\sqrt{x}\cdot\sqrt{x}}=\dfrac{x\sqrt{3x}}{2x}=\dfrac{\sqrt{3x}}{2}\to\boxed{4.}\\\\Used:\\\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{a}\cdot\sqrt{a}=a[/tex]

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