frac{3}{4} left(1+frac{1}{3}right) left(1+fra | vedclass

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(B) The given expression is $\frac{3}{4} 	imes (1+\frac{1}{3}) 	imes (1+\frac{2}{3}) 	imes (1-\frac{2}{5}) 	imes (1+\frac{6}{7}) 	imes (1-\frac{1

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$\frac{1}{7}$

Vedclass · Answer

(B) The given expression is $\frac{3}{4} 	imes (1+\frac{1}{3}) 	imes (1+\frac{2}{3}) 	imes (1-\frac{2}{5}) 	imes (1+\frac{6}{7}) 	imes (1-\frac{12}{13})$.First,simplify each term inside the parentheses:$(1+\frac{1}{3}) = \frac{4}{3}$$(1+\frac{2}{3}) = \frac{5}{3}$$(1-\frac{2}{5}) = \frac{3}{5}$$(1+\frac{6}{7}) = \frac{13}{7}$$(1-\frac{12}{13}) = \frac{1}{13}$Now,substitute these values back into the expression:$= \frac{3}{4} 	imes \frac{4}{3} 	imes \frac{5}{3} 	imes \frac{3}{5} 	imes \frac{13}{7} 	imes \frac{1}{13}$Cancel out the common terms in the numerator and denominator:$= (\frac{3}{4} 	imes \frac{4}{3}) 	imes (\frac{5}{3} 	imes \frac{3}{5}) 	imes (\frac{13}{7} 	imes \frac{1}{13})$$= 1 	imes 1 	imes \frac{1}{7} = \frac{1}{7}$.

$\frac{3}{4} \left(1+\frac{1}{3}\right) \left(1+\frac{2}{3}\right) \left(1-\frac{2}{5}\right) \left(1+\frac{6}{7}\right) \left(1-\frac{12}{13}\right)$ is equal to

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