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The best approach here is to first rationalise the denominator. We do this by multiplying both top and bottom by an approriate value.

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For example to rationalise the denominator for an expression like $\\frac{a}{\\sqrt{b}}$, we multiply numerator and denominator by $\\sqrt{b}$ to get $\\frac{a\\sqrt{b}}{b}$

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Similarly to rationalise the denominator for an expression like $\\frac{a+\\sqrt{b}}{c+\\sqrt{d}}$, we multiply numerator and denominator by $c-\\sqrt{d}$ to get $\\frac{(a+\\sqrt{b})(c-\\sqrt{d})}{c^2-d}$

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Dont forget that:

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$\\sqrt{a \\times b} = \\sqrt{a} \\times \\sqrt{b}$

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$\\sqrt{\\frac{a}{b}} = \\frac{\\sqrt{a}}{\\sqrt{b}}$

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$(a+\\sqrt{b})(a-\\sqrt{b})=a^2-b$

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Remember to check that your answer is in its simplest form.

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$\\frac{\\var{a1}}{\\sqrt{\\var{a2}}}=$[[0]]$\\sqrt{}$[[1]][[2]]

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\n

$\\frac{\\var{b1}}{\\sqrt{\\var{b2}}}=$[[0]]$\\sqrt{}$[[1]][[2]]

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$\\frac{\\var{c1}}{\\sqrt{\\var{c2}}}=$[[0]]$\\sqrt{}$[[1]][[2]]

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$\\frac{\\var{d4}\\sqrt{\\var{d1}}}{\\sqrt{\\var{d2}}}=$[[0]]$\\sqrt{}$[[1]][[2]]

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$\\frac{\\var{e1}\\sqrt{\\var{e2}}}{\\var{e3}\\sqrt{\\var{e4}}}=$[[0]]$\\sqrt{}$[[1]][[2]]

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$\\frac{\\var{f1}\\sqrt{\\var{f6}}}{\\var{f2}\\sqrt{\\var{f5}}}=$[[0]]$\\sqrt{}$[[1]][[2]]

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Express each of the following in the form $\\frac{a\\sqrt{b}}{c}$ where a, b and c are integers, where a and c have no common factors

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