Question

√294-5√3/2+√252-3√1/6. Simplify

Answer 1

Answer:

Hence, the value is $6\sqrt{7} +4\sqrt{6}$.

Step-by-step explanation:

'Prime factorization of any number means to represent that number as a product of prime numbers.'

Split the given number into prime factors.

Form pairs of similar factors such that both factors in each pair should be equal

Take one factor from the pair, split into product of various number

If same number comes in twice, then take out one number from square root and single numbers should be under the square root. The product of this should be equal to the given number.

Given that $\sqrt{294} -5\sqrt{\frac{3}{2} } +\sqrt{252} -3\sqrt{\frac{1}{6} }$

We have to simplify the above one

Square root of $\sqrt{294}$

$\sqrt{294} = \sqrt{2*3*7*7} \\ \sqrt{294} = 7\sqrt{6}$

Square root of $\sqrt{252$

$\sqrt{252} = \sqrt{6*6*7} \\\sqrt{252} =6\sqrt{7}$

now simplify the given question, then

$\sqrt{294} -5\sqrt{\frac{3}{2} } +\sqrt{252} -3\sqrt{\frac{1}{6} }$

=$7\sqrt{6} -5*\frac{\sqrt{3}*\sqrt{2} }{\sqrt{2} \sqrt{2} } +6\sqrt{7} -3*\frac{1*\sqrt{6} }{\sqrt{6} *\sqrt{6} }$

=$7\sqrt{6} -\frac{5}{2} *\sqrt{6} +6\sqrt{7} -\frac{1}{2} *\sqrt{6}$

=$6\sqrt{7} +4\sqrt{6}$

Therefore the answer is $6\sqrt{7} +4\sqrt{6}$

Answer 2

Answer:

please see the above attachment 4 ur ans