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Question:
72×9327×343=? \frac {7^2 \times 9^3}{27 \times 343} = ?27×34372×93=?
A$\frac {27}{7}$
B$\frac {27}{21}$
C$\frac {7}{27}$
D$\frac {21}{27}$
Answer:
72×9327×343 \frac {7^2 \times 9^3}{27 \times 343} 27×34372×93
=72×9333×73= \frac {7^2 \times 9^3}{3^3 \times 7^3} =33×7372×93
=72×(32)333×73= \frac {7^2 \times (3^{2)^3}}{3^3 \times 7^3} =33×7372×(32)3
=72×3633×73= \frac {7^2 \times 3^6}{3^3 \times 7^3}=33×7372×36
=337= \frac {3^3}{7}=733
=27/7= 27/7=27/7
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12\frac{1}{2}21+ 13\frac{1}{3}31+ ________ = 1
