Question

simplify root18/root3+root6-root48/root2+root6+root6/root2+root3​

Answer 1

Step-by-step explanation:

I have done it. Please check it. Hope it will be helpful.

Answer 2

Simplest form of the given data is $\sqrt{6}$ + 2$\sqrt{3}$.

To find :

${\frac{\sqrt{18} }{\sqrt{3} } } + \sqrt{6} - \frac{\sqrt{48} }{\sqrt{2} }+\sqrt{6} +\frac{\sqrt{6} }{\sqrt{2} } +\sqrt{3}$

Given :

${\frac{\sqrt{18} }{\sqrt{3} } } + \sqrt{6} - \frac{\sqrt{48} }{\sqrt{2} }+\sqrt{6} +\frac{\sqrt{6} }{\sqrt{2} } +\sqrt{3}$

Then we have to simplify by using BODMAS rule.

BODMAS RULE

BODMAS

Where the letter "B" stands for Bracket

The letter "O" stands for of.

The letter "D" stands for Division.

The letter "M" stands for Multiplication.

The letter "A" stands for Addition.

The letter "S" stands for Subtraction.

By applying the above rule the problems are solved.

Solution:

Simplifying the given data, we get

${\frac{\sqrt{18} }{\sqrt{3} } } + \sqrt{6} - \frac{\sqrt{48} }{\sqrt{2} }+\sqrt{6} +\frac{\sqrt{6} }{\sqrt{2} } +\sqrt{3}$ $= \sqrt{6} +\sqrt{6} - \sqrt{24} +\sqrt{6} +\sqrt{3} +\sqrt{3}$

From above resultant there are 3 terms of $\sqrt{6}$ and 2 terms of $\sqrt{3}$.

It can be written as,

$\sqrt{6} +\sqrt{6} - \sqrt{24} +\sqrt{6} +\sqrt{3} +\sqrt{3}= 3\times\sqrt{6} +2\times\sqrt{3}-\sqrt{24}$

"$\sqrt{24}$" can be written as "$2\times\sqrt{6}$".

$3\times\sqrt{6} +2\times\sqrt{3}-\sqrt{24} = 3\times\sqrt{6} +2\times\sqrt{3} - 2\times\sqrt{6}$

Subtracting the like terms we gt,

$3\times\sqrt{6} +2\times\sqrt{3} - 2\times\sqrt{6} = \sqrt{6} +2\times\sqrt{3}$                                            

Therefore, simplifying the given data the value is $\sqrt{6} +2\times\sqrt{3}$.

         

To learn more

1) Simplify

5 \sqrt[6]{12}  \div  \sqrt{3}  \times  \sqrt[3]{2} The value of

\sqrt[3]{1 +  \sqrt{2} }  \times  \sqrt[6]{3 - 2 \sqrt{2}  }  is equal to

brainly.in/question/13322697

2) Simplify

5 \sqrt[6]{12}  \div  \sqrt{3}  \times  \sqrt[3]{2}

brainly.in/question/10289120