Question

Hi guys! Could you help me with this exercises? How I should solve it : {
`√(2)` } + {
`√(3)` } + {
`√(2)` +
`√(3)` } ? Please help me, thank you.

Answer 1

1 Rationalize the denominator: \frac{3\sqrt{2}}{\sqrt{6}-\sqrt{3}}(\frac{\sqrt{6}+\sqrt{3}}{\sqrt{6}+\sqrt{3}})​√​6​​​−√​3​​​​​3√​2​​​​​(​√​6​​​+√​3​​​​​√​6​​​+√​3​​​​​)
\frac{6\sqrt{3}+3\sqrt{6}}{{\sqrt{6}}^{2}-{\sqrt{3}}^{2}}-\frac{3}{3-\sqrt{6}}​√​6​​​​2​​−√​3​​​​2​​​​6√​3​​​+3√​6​​​​​−​3−√​6​​​​​3​​
2 Factor out the common term 33
\frac{3(2\sqrt{3}+\sqrt{6})}{{\sqrt{6}}^{2}-{\sqrt{3}}^{2}}-\frac{3}{3-\sqrt{6}}​√​6​​​​2​​−√​3​​​​2​​​​3(2√​3​​​+√​6​​​)​​−​3−√​6​​​​​3​​
3 Use this rule: {\sqrt{x}}^{2}=x√​x​​​​2​​=x
\frac{3(2\sqrt{3}+\sqrt{6})}{6-{\sqrt{3}}^{2}}-\frac{3}{3-\sqrt{6}}​6−√​3​​​​2​​​​3(2√​3​​​+√​6​​​)​​−​3−√​6​​​​​3​​
4 Use this rule: {\sqrt{x}}^{2}=x√​x​​​​2​​=x
\frac{3(2\sqrt{3}+\sqrt{6})}{6-3}-\frac{3}{3-\sqrt{6}}​6−3​​3(2√​3​​​+√​6​​​)​​−​3−√​6​​​​​3​​
5 Simplify 6-36−3 to 33
\frac{3(2\sqrt{3}+\sqrt{6})}{3}-\frac{3}{3-\sqrt{6}}​3​​3(2√​3​​​+√​6​​​)​​−​3−√​6​​​​​3​​
6 Cancel 33
2\sqrt{3}+\sqrt{6}-\frac{3}{3-\sqrt{6}}2√​3​​​+√​6​​​−​3−√​6​​​​​3​​
7 Rationalize the denominator: \frac{3}{3-\sqrt{6}}(\frac{3+\sqrt{6}}{3+\sqrt{6}})​3−√​6​​​​​3​​(​3+√​6​​​​​3+√​6​​​​​)