1/3-1/7-1/13/2/3-2/7-2/3×3/4-3/16-3/64/1-1/4-1/16-1/64+5/8
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\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{264}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)
\(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\cdot\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{264}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}\cdot\frac{\frac{3}{4}\left(1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}\right)}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)\(+\frac{5}{8}\)
\(\frac{1}{2}\cdot\frac{3}{4}+\frac{5}{8}=\frac{3}{8}+\frac{5}{8}=1\)
\(B=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
=>\(B=\frac{1.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}{3.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{14}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{\frac{4}{4}-\frac{4}{16}-\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)
=>\(B=\frac{1}{3}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)
=>\(B=\frac{1}{3}.\frac{3}{4}+\frac{5}{8}\)
=>\(B=\frac{1}{4}+\frac{5}{8}\)
=>\(B=\frac{2}{8}+\frac{5}{8}\)
=>\(B=\frac{7}{8}\)
l-i-k-e cho mình nhé bạn.
Sửa đề
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}\right)}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot3+\dfrac{5}{8}=\dfrac{3}{2}+\dfrac{5}{8}=\dfrac{17}{8}\)
A= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{\dfrac{4}{4}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{4^3}-\dfrac{1}{16^2})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{4^2}-\dfrac{1}{16^2})}{4-\dfrac{1}{4^3}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{16^2})}{4.-\dfrac{1}{4^2}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{264}\right)}{\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{264}}\)
\(=\dfrac{1}{2}.3=\dfrac{3}{2}\)