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To solve the expression [tex]\(\sqrt{2 \frac{13}{16}} - \frac{2}{\sqrt{5}}\)[/tex], we will break it down into smaller parts and then combine them.
1. Rewrite the mixed number [tex]\(2 \frac{13}{16}\)[/tex] as an improper fraction:
We have:
[tex]\[ 2 \frac{13}{16} = 2 + \frac{13}{16} \][/tex]
To convert [tex]\(2\)[/tex] to a fraction with a denominator of [tex]\(16\)[/tex]:
[tex]\[ 2 = \frac{2 \cdot 16}{16} = \frac{32}{16} \][/tex]
Therefore:
[tex]\[ 2 \frac{13}{16} = \frac{32}{16} + \frac{13}{16} = \frac{32 + 13}{16} = \frac{45}{16} \][/tex]
2. Calculate the square root of [tex]\(\frac{45}{16}\)[/tex]:
[tex]\[ \sqrt{\frac{45}{16}} = \frac{\sqrt{45}}{\sqrt{16}} = \frac{\sqrt{45}}{4} \][/tex]
We can simplify [tex]\(\sqrt{45}\)[/tex]:
[tex]\[ \sqrt{45} = \sqrt{9 \cdot 5} = \sqrt{9} \cdot \sqrt{5} = 3\sqrt{5} \][/tex]
Thus:
[tex]\[ \sqrt{\frac{45}{16}} = \frac{3\sqrt{5}}{4} \][/tex]
3. Rewrite [tex]\(\frac{2}{\sqrt{5}}\)[/tex] by rationalizing the denominator:
[tex]\[ \frac{2}{\sqrt{5}} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \][/tex]
4. Subtract the two terms [tex]\(\frac{3\sqrt{5}}{4}\)[/tex] and [tex]\(\frac{2\sqrt{5}}{5}\)[/tex] by getting a common denominator:
The common denominator of [tex]\(4\)[/tex] and [tex]\(5\)[/tex] is [tex]\(20\)[/tex]:
Rewrite the first term:
[tex]\[ \frac{3\sqrt{5}}{4} = \frac{3\sqrt{5} \cdot 5}{4 \cdot 5} = \frac{15\sqrt{5}}{20} \][/tex]
Rewrite the second term:
[tex]\[ \frac{2\sqrt{5}}{5} = \frac{2\sqrt{5} \cdot 4}{5 \cdot 4} = \frac{8\sqrt{5}}{20} \][/tex]
Now subtract the two fractions:
[tex]\[ \frac{15\sqrt{5}}{20} - \frac{8\sqrt{5}}{20} = \frac{15\sqrt{5} - 8\sqrt{5}}{20} = \frac{(15 - 8)\sqrt{5}}{20} = \frac{7\sqrt{5}}{20} \][/tex]
Thus, the answer in the form [tex]\(\frac{a \sqrt{5}}{b}\)[/tex] is:
[tex]\[ \frac{7\sqrt{5}}{20} \][/tex]
where [tex]\(a = 7\)[/tex] and [tex]\(b = 20\)[/tex].
1. Rewrite the mixed number [tex]\(2 \frac{13}{16}\)[/tex] as an improper fraction:
We have:
[tex]\[ 2 \frac{13}{16} = 2 + \frac{13}{16} \][/tex]
To convert [tex]\(2\)[/tex] to a fraction with a denominator of [tex]\(16\)[/tex]:
[tex]\[ 2 = \frac{2 \cdot 16}{16} = \frac{32}{16} \][/tex]
Therefore:
[tex]\[ 2 \frac{13}{16} = \frac{32}{16} + \frac{13}{16} = \frac{32 + 13}{16} = \frac{45}{16} \][/tex]
2. Calculate the square root of [tex]\(\frac{45}{16}\)[/tex]:
[tex]\[ \sqrt{\frac{45}{16}} = \frac{\sqrt{45}}{\sqrt{16}} = \frac{\sqrt{45}}{4} \][/tex]
We can simplify [tex]\(\sqrt{45}\)[/tex]:
[tex]\[ \sqrt{45} = \sqrt{9 \cdot 5} = \sqrt{9} \cdot \sqrt{5} = 3\sqrt{5} \][/tex]
Thus:
[tex]\[ \sqrt{\frac{45}{16}} = \frac{3\sqrt{5}}{4} \][/tex]
3. Rewrite [tex]\(\frac{2}{\sqrt{5}}\)[/tex] by rationalizing the denominator:
[tex]\[ \frac{2}{\sqrt{5}} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \][/tex]
4. Subtract the two terms [tex]\(\frac{3\sqrt{5}}{4}\)[/tex] and [tex]\(\frac{2\sqrt{5}}{5}\)[/tex] by getting a common denominator:
The common denominator of [tex]\(4\)[/tex] and [tex]\(5\)[/tex] is [tex]\(20\)[/tex]:
Rewrite the first term:
[tex]\[ \frac{3\sqrt{5}}{4} = \frac{3\sqrt{5} \cdot 5}{4 \cdot 5} = \frac{15\sqrt{5}}{20} \][/tex]
Rewrite the second term:
[tex]\[ \frac{2\sqrt{5}}{5} = \frac{2\sqrt{5} \cdot 4}{5 \cdot 4} = \frac{8\sqrt{5}}{20} \][/tex]
Now subtract the two fractions:
[tex]\[ \frac{15\sqrt{5}}{20} - \frac{8\sqrt{5}}{20} = \frac{15\sqrt{5} - 8\sqrt{5}}{20} = \frac{(15 - 8)\sqrt{5}}{20} = \frac{7\sqrt{5}}{20} \][/tex]
Thus, the answer in the form [tex]\(\frac{a \sqrt{5}}{b}\)[/tex] is:
[tex]\[ \frac{7\sqrt{5}}{20} \][/tex]
where [tex]\(a = 7\)[/tex] and [tex]\(b = 20\)[/tex].
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