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reshav

· started a discussion

· 1 Months ago

72 (Question:110057.0)

is there any other shortcut method to do this

Question:
The square root of \(\cfrac{\left ( 3\cfrac {1}{4} \right )^4-\left ( 4\cfrac {1}{3} \right )^4}{\left ( 3\cfrac {1}{4} \right )^2-\left ( 4\cfrac {1}{3} \right )^2} \)   is:
Options:
A) \(5\cfrac{1}{12}\)
B) \(5\cfrac{4}{13}\)
C) \(5\cfrac{5}{12}\)
D) \(5\cfrac{9}{13}\)
Solution:
Ans: (c) Let a = \(\left (3 \cfrac {1}{4} \right )^2\) and b = \(\left (4 \cfrac {1}{3} \right )^2\)

\(\therefore\) Expression = \(\cfrac{a^2-b^2}{a-b}= a+b\)

\(\cfrac{\left (3 \cfrac {1}{4} \right )^4-\left ( 4\cfrac {1}{3} \right )^4}{{\left (3 \cfrac {1}{4} \right )^2-\left ( 4\cfrac {1}{3} \right )^2}}\)

\(\left ( \cfrac {13}{4} \right )^2+\left ( \cfrac {13}{3} \right )^2\)

= 169\(\left ( \cfrac {1}{16} +\cfrac{1}{9} \right )\)

= 169 \(\times\cfrac{25}{144}\)

Required square root = \(\sqrt{169\times\frac{25}{144}}=13 \times\cfrac{5}{12}\)

= \(\cfrac{65}{12} = 5\cfrac{5}{12}\)

x3 = 3775

x = \(\sqrt[3]{3775}\) 

= \(\sqrt[3]{15 \times15\times15} = 15\)

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