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10 Aug 2023, 06:00
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E
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Difficulty:
25%
(medium)
Question Stats:
81% (01:48) correct
19%
(02:31)
wrong
based on 1044
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Which of the following is the value of \(\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2} } + \frac{1}{2 + \sqrt{3} }\)?
A. \(\frac{1}{3 + 2\sqrt{2} + 2\sqrt{3} }\)
B. \(\frac{3}{3 + 2\sqrt{2} + 2\sqrt{3} }\)
C. 1/3
D. 3/7
E. 1
A. \(\frac{1}{3 + 2\sqrt{2} + 2\sqrt{3} }\)
B. \(\frac{3}{3 + 2\sqrt{2} + 2\sqrt{3} }\)
C. 1/3
D. 3/7
E. 1
10 Aug 2023, 08:56
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Bunuel
The square root in the denominator should get you thinking on how to remove them.
\(\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2} } + \frac{1}{2 + \sqrt{3} }\)
\(\frac{1*(\sqrt{2} - 1)}{(\sqrt{2} +1)(\sqrt{2} - 1)} + \frac{1*(\sqrt{3} - \sqrt{2} )}{(\sqrt{3} + \sqrt{2}) (\sqrt{3} - \sqrt{2} ) } + \frac{1 (2 - \sqrt{3}) }{(2 + \sqrt{3}) (2 - \sqrt{3}) }\)
\(\frac{(\sqrt{2} - 1)}{2-1} + \frac{(\sqrt{3} - \sqrt{2} )}{3-2 } + \frac{(2 - \sqrt{3}) }{4-3 }\)
\({(\sqrt{2} - 1)}+ {(\sqrt{3} - \sqrt{2} )}+ {(2 - \sqrt{3}) }\)
\(2-1\)
\(1\)
E
You could also guess
\(\frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{3} + \sqrt{2} } + \frac{1}{2 + \sqrt{3} }\)
\(\frac{1}{1.4 + 1} + \frac{1}{1.7 + 1.4} + \frac{1}{2+1.7}\)
\(\frac{1}{2.4 } + \frac{1}{3.1} + \frac{1}{3.7}\)
The first fraction is more than 1/3 just less than 1/2, second term is around 1/3. Just these two combined should give you answer greater than 2/3.
Only E is left.
15 Aug 2023, 14:10
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Bunuel
We can make the denominators “disappear” if we remove the radicals from the denominators.
How can we remove those radicals without changing the values of the fractions? By multiplying each fraction by a corresponding fraction that is equal to one:
(1/[√2+1]) × (√2–1)/(√2–1) = (√2–1)/(2–1) = √2 – 1
(1/[√3+√2]) × (√3–√2)/(√3–√2) = (√3–√2)/(3–2) = √3 – √2
(1/[2+√3]) × (2–√3)/(2–√3) = (2–√3)/(4–3) = 2 – √3
So, we have:
√2 – 1 + √3 – √2 + 2 – √3 = 1
Answer: E
