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\frac{\left(4+\sqrt{5}\right)\left(4+\sqrt{5}\right)}{\left(4-\sqrt{5}\right)\left(4+\sqrt{5}\right)}+\frac{4-\sqrt{5}}{4+\sqrt{5}}
Rationalize the denominator of \frac{4+\sqrt{5}}{4-\sqrt{5}} by multiplying numerator and denominator by 4+\sqrt{5}.
\frac{\left(4+\sqrt{5}\right)\left(4+\sqrt{5}\right)}{4^{2}-\left(\sqrt{5}\right)^{2}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}
Consider \left(4-\sqrt{5}\right)\left(4+\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{\left(4+\sqrt{5}\right)\left(4+\sqrt{5}\right)}{16-5}+\frac{4-\sqrt{5}}{4+\sqrt{5}}
Square 4. Square \sqrt{5}.
\frac{\left(4+\sqrt{5}\right)\left(4+\sqrt{5}\right)}{11}+\frac{4-\sqrt{5}}{4+\sqrt{5}}
Subtract 5 from 16 to get 11.
\frac{\left(4+\sqrt{5}\right)^{2}}{11}+\frac{4-\sqrt{5}}{4+\sqrt{5}}
Multiply 4+\sqrt{5} and 4+\sqrt{5} to get \left(4+\sqrt{5}\right)^{2}.
\frac{16+8\sqrt{5}+\left(\sqrt{5}\right)^{2}}{11}+\frac{4-\sqrt{5}}{4+\sqrt{5}}
Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+\sqrt{5}\right)^{2}.
\frac{16+8\sqrt{5}+5}{11}+\frac{4-\sqrt{5}}{4+\sqrt{5}}
The square of \sqrt{5} is 5.
\frac{21+8\sqrt{5}}{11}+\frac{4-\sqrt{5}}{4+\sqrt{5}}
Add 16 and 5 to get 21.
\frac{21+8\sqrt{5}}{11}+\frac{\left(4-\sqrt{5}\right)\left(4-\sqrt{5}\right)}{\left(4+\sqrt{5}\right)\left(4-\sqrt{5}\right)}
Rationalize the denominator of \frac{4-\sqrt{5}}{4+\sqrt{5}} by multiplying numerator and denominator by 4-\sqrt{5}.
\frac{21+8\sqrt{5}}{11}+\frac{\left(4-\sqrt{5}\right)\left(4-\sqrt{5}\right)}{4^{2}-\left(\sqrt{5}\right)^{2}}
Consider \left(4+\sqrt{5}\right)\left(4-\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{21+8\sqrt{5}}{11}+\frac{\left(4-\sqrt{5}\right)\left(4-\sqrt{5}\right)}{16-5}
Square 4. Square \sqrt{5}.
\frac{21+8\sqrt{5}}{11}+\frac{\left(4-\sqrt{5}\right)\left(4-\sqrt{5}\right)}{11}
Subtract 5 from 16 to get 11.
\frac{21+8\sqrt{5}}{11}+\frac{\left(4-\sqrt{5}\right)^{2}}{11}
Multiply 4-\sqrt{5} and 4-\sqrt{5} to get \left(4-\sqrt{5}\right)^{2}.
\frac{21+8\sqrt{5}}{11}+\frac{16-8\sqrt{5}+\left(\sqrt{5}\right)^{2}}{11}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-\sqrt{5}\right)^{2}.
\frac{21+8\sqrt{5}}{11}+\frac{16-8\sqrt{5}+5}{11}
The square of \sqrt{5} is 5.
\frac{21+8\sqrt{5}}{11}+\frac{21-8\sqrt{5}}{11}
Add 16 and 5 to get 21.
\frac{21+8\sqrt{5}+21-8\sqrt{5}}{11}
Since \frac{21+8\sqrt{5}}{11} and \frac{21-8\sqrt{5}}{11} have the same denominator, add them by adding their numerators.
\frac{42}{11}
Do the calculations in 21+8\sqrt{5}+21-8\sqrt{5}.