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Mathematics

Evaluate:

(7−1−8−1)−(3−1−4−1)−1(7^{-1} - 8^{-1}) - (3^{-1} - 4^{-1})^{-1}

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Answer

(7−1−8−1)−(3−1−4−1)−1=(17−18)−(13−14)−1(7^{-1} - 8^{-1}) - (3^{-1} - 4^{-1})^{-1}\\[1em] = \Big(\dfrac{1}{7} - \dfrac{1}{8}\Big) - \Big(\dfrac{1}{3} - \dfrac{1}{4}\Big)^{-1}

LCM of 7 and 8 is 2 x 2 x 2 x 7 = 56

And

LCM of 3 and 4 is 2 x 2 x 3 = 12

=(1×87×8−1×78×7)−(1×43×4−1×34×3)−1=(856−756)−(412−312)−1=(8−756)−(4−312)−1=(156)−(112)−1=(156)−(121)1= \Big(\dfrac{1 \times 8}{7 \times 8} - \dfrac{1 \times 7}{8 \times 7}\Big) - \Big(\dfrac{1 \times 4}{3 \times 4} - \dfrac{1 \times 3}{4 \times 3}\Big)^{-1}\\[1em] = \Big(\dfrac{8}{56} - \dfrac{7}{56}\Big) - \Big(\dfrac{4}{12} - \dfrac{3}{12}\Big)^{-1}\\[1em] = \Big(\dfrac{8 - 7}{56}\Big) - \Big(\dfrac{4 - 3}{12}\Big)^{-1}\\[1em] = \Big(\dfrac{1}{56}\Big) - \Big(\dfrac{1}{12}\Big)^{-1}\\[1em] = \Big(\dfrac{1}{56}\Big) - \Big(\dfrac{12}{1}\Big)^1

LCM of 56 and 1 is 56

=(156)−(12×561×56)=(156)−(67256)=(1−67256)=−67156=−115556= \Big(\dfrac{1}{56}\Big) - \Big(\dfrac{12 \times 56}{1 \times 56}\Big)\\[1em] = \Big(\dfrac{1}{56}\Big) - \Big(\dfrac{672}{56}\Big)\\[1em] = \Big(\dfrac{1 - 672}{56}\Big)\\[1em] = -\dfrac{671}{56}\\[1em] = -11\dfrac{55}{56}

(7−1−8−1)−(3−1−4−1)−1=−115556(7^{-1} - 8^{-1}) - (3^{-1} - 4^{-1})^{-1} = -11\dfrac{55}{56}

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