Question

rationalize and denominator
16 /√41-5​

Answer 1

Answer:

• √41 + 5 is the answer

Step-by-step explanation:

${\small{\tt{\dfrac{16}{\sqrt{41} - 5}}}} \\ \\ {\small{ \hookrightarrow{ \tt{ \dfrac{16}{\sqrt{41} - 5}}}} \times \dfrac{\sqrt{41} + 5}{\sqrt{41} + 5}} \\ \\{ \small{ \hookrightarrow{ \tt{ \dfrac{16 \sqrt{41} + 80}{ ({ \sqrt{41}) }^{2} - {5}^{2} }}}}} \\ \\{ \small{ \hookrightarrow{ \tt{ \dfrac{16 \sqrt{41} + 80}{41 - 25}}}}} \\ \\ { \small{ \hookrightarrow{ \tt{ \dfrac{16 \sqrt{41} + 80}{16}}}}} \\ \\ { \small{ \hookrightarrow{ \tt{ \dfrac{{ \cancel{16}} (\sqrt{41} + 5)}{{ \cancel{16}}}}}}} \\ \\ { \small{ \hookrightarrow{{ \tt{ \sqrt{41} + 5 }}}}} \\ \\ { \circ \: { \small{ \underline{ \boxed{ \frak{ \purple{\underline{ \sqrt{41} + 5} }\: is \: the \: required \: answer}}}}}}$

Answer 2

Given

16/(root41 - 5)

=16(root41+5)/[(root 41 -5)(root41+5)]

=[16(root41+5)]/[(root41)^2 -(5)^2]

=[16(root41+5)]/(41-25)

=16(root41+5]/16

=root41+5