Question

1. \\(\frac{5}{2 + \sqrt{3}} = \frac{1 - \sqrt{3}}{x}\\)

Explanation:

Step1: Cross - multiply the equation

Cross - multiplying gives us \(5x=(2 + \sqrt{3})(1-\sqrt{3})\).

Step2: Expand the right - hand side

Using the formula \((a + b)(c + d)=ac+ad+bc+bd\), we have:
\((2+\sqrt{3})(1 - \sqrt{3})=2\times1-2\times\sqrt{3}+\sqrt{3}\times1-\sqrt{3}\times\sqrt{3}\)
\(=2-2\sqrt{3}+\sqrt{3}-3\)
\(=(2 - 3)+(-2\sqrt{3}+\sqrt{3})\)
\(=-1-\sqrt{3}\)

Step3: Solve for \(x\)

From \(5x=-1 - \sqrt{3}\), we get \(x=\frac{-1-\sqrt{3}}{5}\) or we can rationalize the denominator of the original fraction first. Let's try rationalizing the left - hand side fraction \(\frac{5}{2+\sqrt{3}}\) by multiplying numerator and denominator by \(2-\sqrt{3}\):
\(\frac{5(2 - \sqrt{3})}{(2+\sqrt{3})(2 - \sqrt{3})}=\frac{10-5\sqrt{3}}{4-3}=10 - 5\sqrt{3}\)
Now our equation is \(10 - 5\sqrt{3}=\frac{1-\sqrt{3}}{x}\)
Cross - multiplying gives \(x(10 - 5\sqrt{3})=1-\sqrt{3}\)
Factor out 5 from the left - hand side: \(5x(2-\sqrt{3})=1 - \sqrt{3}\)
Then \(x=\frac{1-\sqrt{3}}{5(2 - \sqrt{3})}\)
Multiply numerator and denominator by \(2+\sqrt{3}\):
\(x=\frac{(1 - \sqrt{3})(2+\sqrt{3})}{5(2 - \sqrt{3})(2+\sqrt{3})}\)
Expand the numerator: \((1 - \sqrt{3})(2+\sqrt{3})=2+\sqrt{3}-2\sqrt{3}-3=-1-\sqrt{3}\)
Denominator: \(5(4 - 3)=5\)
So \(x=\frac{-1-\sqrt{3}}{5}\)

Answer:

\(x = \frac{-1-\sqrt{3}}{5}\)