Skip to main content
Evaluate
Tick mark Image

Similar Problems from Web Search

Share

\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)
Rationalize the denominator of \frac{1}{\sqrt{2}-1} by multiplying numerator and denominator by \sqrt{2}+1.
\frac{\sqrt{2}+1}{\left(\sqrt{2}\right)^{2}-1^{2}}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)
Consider \left(\sqrt{2}-1\right)\left(\sqrt{2}+1\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{\sqrt{2}+1}{2-1}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)
Square \sqrt{2}. Square 1.
\frac{\sqrt{2}+1}{1}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)
Subtract 1 from 2 to get 1.
\sqrt{2}+1+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)
Anything divided by one gives itself.
\sqrt{2}+1+\left(\sqrt{3}\right)^{2}-\sqrt{3}\sqrt{6}
Use the distributive property to multiply \sqrt{3} by \sqrt{3}-\sqrt{6}.
\sqrt{2}+1+3-\sqrt{3}\sqrt{6}
The square of \sqrt{3} is 3.
\sqrt{2}+1+3-\sqrt{3}\sqrt{3}\sqrt{2}
Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.
\sqrt{2}+1+3-3\sqrt{2}
Multiply \sqrt{3} and \sqrt{3} to get 3.
\sqrt{2}+4-3\sqrt{2}
Add 1 and 3 to get 4.
-2\sqrt{2}+4
Combine \sqrt{2} and -3\sqrt{2} to get -2\sqrt{2}.