Question

4+√6/√8ratinalize denominetar​

Answer 1

We are asked to rationalise the denominator:

$:\implies \sf \dfrac{4+\sqrt{6}}{\sqrt{8}} \\ \\ :\implies \sf \dfrac{4+\sqrt{6}}{\sqrt{8}} \times \dfrac{\sqrt{8}}{\sqrt{8}} \\ \\ :\implies \sf \dfrac{4 \sqrt{8} + \sqrt{48}}{8}$

Henceforth, solved!

Some identities:

$\; \; \; \; \; \; \;{\sf{\leadsto (a+b)^{2} \: = \: = a^{2} \: + \: 2ab \: + b^{2}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (a-b)^{2} \: = \: a^{2} \: - \: 2ab \: + b^{2}}}$

$\; \; \; \; \; \; \;{\sf{\leadsto a^{2} \: - b^{2} \: = \: (a+b) \: (a-b)}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (x+a) (x+b) = x^2 + (a+b)x + ab}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (x+y+z)^{2} \: = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (x+y)^{3} \: = x^3 + y^3 + 3xy (x+y)}}$

$\; \; \; \; \; \; \;{\sf{\leadsto (x-y)^{3} \: = x^3 - y^3 -3xy(x-y) \: = x^3 - 3x^2y + 3xy^2 - y^3}}$

$\; \; \; \; \; \; \;{\sf{\leadsto x^3 + y^3 + z^3 - 3zyx \: = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)}}$