\left(\dfrac{11}8-\dfrac5{16}+\dfrac1{32}\right)\cdot\dfrac8{15}( 8 11 − 16 5 + 32 1 )⋅ 15 8 .

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вопрос

First, let's simplify the expression inside the parentheses:
(\dfrac{11}{8}-\dfrac{5}{16}+\dfrac{1}{32})

To do this, we need to find a common denominator for all the fractions. The least common multiple of 8, 16, and 32 is 32.

Rewriting the fractions with a common denominator of 32:
(\dfrac{11 \cdot 4}{8 \cdot 4} - \dfrac{5 \cdot 2}{16 \cdot 2} + \dfrac{1}{32} \
= \dfrac{44}{32} - \dfrac{10}{32} + \dfrac{1}{32} \
= \dfrac{44 - 10 + 1}{32} \
= \dfrac{35}{32})

Now we multiply by (\dfrac{8}{15}):
(\dfrac{35}{32} \cdot \dfrac{8}{15} \
= \dfrac{35 \cdot 8}{32 \cdot 15} \
= \dfrac{280}{480} \
= \dfrac{7}{12})

Therefore,
((\dfrac{11}{8}-\dfrac{5}{16}+\dfrac{1}{32}) \cdot \dfrac{8}{15} = \dfrac{7}{12})