Question

simplify each of the following by rationalizing the denominator. (7+√5) / (7-√5)​

Answer 1

I think this answer may correct✅

Answer 2

$\: \: \: \: \implies \sf \dfrac{(7+√5) }{ (7-√5)}$

• Rationalizing The Denominator Term

$\\{ \: \: \: \: \implies \sf \dfrac{(7+√5) }{ (7-√5)} \times \dfrac{(7+√5) }{ (7 + √5)}}$

${ \: \: \: \: \implies \sf \dfrac{(7+ \sqrt{5} )^{2} }{ {7}^{2} - ( { \sqrt{5} })^{2} }}$

${ \: \: \: \: \implies \sf \dfrac{ {7}^{2} + { \sqrt{5} }^{2} + 2 \times 7 \times \sqrt{5} }{ {7}^{2} - ( { \sqrt{5} })^{2} }}$

${ \: \: \: \: \implies \sf \dfrac{ 49 + 5 + 14 \sqrt{5} }{ {7}^{2} - ( { \sqrt{5} })^{2} }}$

${ \: \: \: \: \implies \sf \dfrac{ 54 + 14 \sqrt{5} }{ {7}^{2} - 5 }}$

${ \: \: \: \: \implies \sf \dfrac{ 2(27 + 7 \sqrt{5} ) }{ 49 - 5 }}$

${ \: \: \: \: \implies \sf \dfrac{ 2(27 + 7 \sqrt{5} ) }{ 44 }}$

${ \: \: \: \: \implies \sf \dfrac{ 2(27 + 7 \sqrt{5} ) }{ 2(22) }}$

${ \: \: \: \: \implies \sf \dfrac{ 27 + 7 \sqrt{5} }{ 22 }}$

$~~~~~~~~~~~~~~~~~\boxed{{\bf \dfrac{(7+√5) }{ (7-√5)}=\sf \dfrac{ 27 + 7 \sqrt{5} }{ 22 }}}$

• $\\\\\begin{gathered}\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\gray{\begin{gathered}\tiny\begin{gathered}\small{\small{\small{\small{\small{\small{\small{\small{\small{\small{\begin{gathered}\begin{gathered}\begin{gathered}\\\\\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\red{ \bigstar} \: \underline{\bf{\orange{More \: Useful \: Formula}}}\\ {\boxed{\begin{array}{cc}\dashrightarrow \sf(a + b)^{2} = {a}^{2} + {b}^{2} + 2ab \\\\\dashrightarrow \sf(a - b)^{2} = {a}^{2} + {b}^{2} - 2ab \\\\\dashrightarrow \sf(a + b)(a - b) = {a}^{2} - {b}^{2} \\\\\dashrightarrow \sf(a + b) ^{3} = {a}^{3} + b^{3} + 3ab(a + b) \\\\ \dashrightarrow\sf(a - b) ^{3} = {a}^{3} - b^{3} - 3ab(a - b) \\ \\\dashrightarrow\sf a ^{3} + {b}^{3} = (a + b)(a ^{2} + {b}^{2} - ab) \\\\\dashrightarrow \sf a ^{3} - {b}^{3} = (a - b)(a ^{2} + {b}^{2} + ab )\\\\\dashrightarrow \sf{a²+b²=(a+b)²-2ab}\\ \end{array}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}}}}}}}}}}}\end{gathered}\end{gathered}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\\ \end{gathered}$