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

User Cezn
by
8.3k points

1 Answer

7 votes
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}}6−33(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}}​33(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​​​​)
User Ahmad Ajmi
by
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