\(\dfrac{12-\dfrac{12}{7}-\dfrac{12}{289}-\dfrac{12}{85}}{4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}}=\)
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\(=81.\dfrac{12.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4.\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6.\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}.\dfrac{158}{711}\)
\(=81.\dfrac{12}{4}:\dfrac{5}{6}.\dfrac{2}{9}\)
\(=243:\dfrac{5}{6}.\dfrac{2}{9}\)
\(=\dfrac{1458}{5}.\dfrac{2}{9}\)
\(=\dfrac{324}{5}\)
98775 - 32 x 85
=98775 -2720
=96055
67500 - 24 x 236
= 67500 -5664
=61836
568 + 101598 : 287
= 568 +354
=922
6875 + 980 -180
=7855 -180
=7675
\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)
\(=\dfrac{7}{10}-\dfrac{1}{2}\)
= \(\dfrac{1}{5}\)
\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)
\(=\dfrac{8}{11}+\dfrac{2}{11}\)
\(=\dfrac{10}{11}\)
\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}:\dfrac{5}{8}\)
\(=\dfrac{1}{6}x\dfrac{8}{5}\)
\(=\dfrac{8}{30}\)
\(=\dfrac{4}{15}\)
\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)
\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)
\(=\dfrac{5}{12}-\dfrac{1}{6}\)
\(=\dfrac{5}{12}-\dfrac{2}{12}\)
\(=\dfrac{3}{12}=\dfrac{1}{4}\)
\(=\dfrac{4}{11}\left(\dfrac{5}{12}-1+\dfrac{7}{12}\right)=\dfrac{4}{11}\cdot0=0\)
a: \(=\dfrac{-2}{12}-\dfrac{5}{12}+\dfrac{7}{12}=0\)
b: \(=\left(\dfrac{4}{45}-\dfrac{1}{45}+\dfrac{7}{45}+\dfrac{4}{45}-\dfrac{2}{45}-\dfrac{9}{45}\right)=\dfrac{3}{45}=\dfrac{1}{15}\)
a: =(2/7-2/7)(-4/7-5/9)=0
b:
Sửa đề: 9/13*(-12/17)+9/13*29/27
=9/13(-12/17+29/17)
=9/13*17/17=9/13
c: \(=\dfrac{1}{7}\left(4+\dfrac{6}{7}+\dfrac{8}{7}\right)=\dfrac{1}{7}\cdot6=\dfrac{6}{7}\)
d: =7/10(5/7+9/7+8/7+13/7)
=5*7/10=7/2
\(=\dfrac{3.4-\dfrac{3.4}{4}-\dfrac{3.4}{289}-\dfrac{3.4}{85}}{4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}}=\dfrac{3.\left(4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}\right)}{1.\left(4-\dfrac{4}{7}-\dfrac{4}{289}-\dfrac{4}{85}\right)}=\dfrac{3}{1}=3\)
=3