Example

3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b

Question

$$\frac{ \sqrt{ 3+ } \sqrt{ 2 } }{ \sqrt{ 3- } \sqrt{ 2 } } + \frac{ \sqrt{ 3- } \sqrt{ 2 } }{ \sqrt{ 3+ } \sqrt{ 2 } } =a \sqrt{ 3 } +b$$

Answer

a=(sqrt(3+)/sqrt(3-)+sqrt(3-)/sqrt(3+)-b)/sqrt(3)

Solution


Cancel \(\sqrt{2}\).
\[\frac{\sqrt{3+}}{\sqrt{3-}}+\frac{\sqrt{3-}\sqrt{2}}{\sqrt{3+}\sqrt{2}}=a\sqrt{3}+b\]
Cancel \(\sqrt{2}\).
\[\frac{\sqrt{3+}}{\sqrt{3-}}+\frac{\sqrt{3-}}{\sqrt{3+}}=a\sqrt{3}+b\]
Regroup terms.
\[\frac{\sqrt{3+}}{\sqrt{3-}}+\frac{\sqrt{3-}}{\sqrt{3+}}=\sqrt{3}a+b\]
Subtract \(b\) from both sides.
\[\frac{\sqrt{3+}}{\sqrt{3-}}+\frac{\sqrt{3-}}{\sqrt{3+}}-b=\sqrt{3}a\]
Divide both sides by \(\sqrt{3}\).
\[\frac{\frac{\sqrt{3+}}{\sqrt{3-}}+\frac{\sqrt{3-}}{\sqrt{3+}}-b}{\sqrt{3}}=a\]
Switch sides.
\[a=\frac{\frac{\sqrt{3+}}{\sqrt{3-}}+\frac{\sqrt{3-}}{\sqrt{3+}}-b}{\sqrt{3}}\]
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