Question

find root 2/6

pls tell the answer​

Answer 1

Answer:

1/(root 3)

Step-by-step explanation:

2/6=1/3

hope it helps

Answer 2

Question :-

$\sf Find\: :\sqrt{\dfrac{2}{6} }$

Answer :-

$\longrightarrow \sf \sqrt{\dfrac{2}{6} }$

Reducing to the lowest terms,

$\implies \sf \sqrt{\dfrac{1}{3} }$

We know that,

$\sf \sqrt[n]{1} = 1$ and $\sf 1^{n} = 1$

Hence, The number obtained is

$\implies \sf \dfrac{1}{\sqrt{3} }$

The denominator is irrational.

By multiplying the numerator and the denominator by √3 in order to rationalize the denominator, we get :

$\implies\sf \dfrac{1}{\sqrt{3} } \times \dfrac{\sqrt{3} }{\sqrt{3} }$

$\implies \sf \dfrac{1 \times \sqrt{3} }{\sqrt{3} \times \sqrt{3} }$

$\implies \sf \dfrac{\sqrt{3} }{(\sqrt{3} )^{2}}$

$\implies \sf \dfrac{\sqrt{3} }{3}$

$\longrightarrow \sf Therefore\:\: \sqrt{ \dfrac{2}{6}} = \dfrac{\sqrt{3} }{3}$

Important points to note :-

$\tiny{\boxed{\begin{array}{cc} \sf (a + b)^{2}  = a^{2} + 2ab + b^{2} \\\sf (a - b)^{2} = a^{2} - 2ab + b^{2} \\ \sf a^2 - b^2 = (a - b)(a + b) \\  \sf(x + a)(x + b) = x^2 + x(a + b) + ab \\ \sf (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)  \\ \sf (a + b)^3 = a^3 + b^3 + 3ab(a + b)  \\ \sf (a - b)^3 = a^3 - b^3 - 3ab(a - b)  \\ \sf a^3 + b^3 = (a + b)(a^2 - ab + b^2)  \\  \sf a^3 - b^3 = (a - b)(a^2 + ab + b^2) \\  \sf a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \\ \sf a^n \times a^b = a^{n+b}  \\ \sf a^n \times b^n= ab^n \\  \sf a^{-n} = 1/a^n  \\ \sf a^n \div a^b = a^{n-b}  \\  \sf (a^n/b^n) = (a/b)^n  \\\sf  a^0 = 1 \\\end{array}}}$