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Trả lời :
\(\frac{56^2}{28^2}=\frac{\left(2.28\right)^2}{28^2}=\frac{2^2.28^2}{28^2}=2^2=4\)
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\(A=\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}\)
\(=\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{7}-\frac{1}{13}\)
\(=\frac{6}{91}\)
\(A=\frac{1}{7\times8}+\frac{1}{8\times9}+\frac{1}{9\times10}+...+\frac{1}{13\times14}\)
\(=\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{13}-\frac{1}{14}\)
\(=\frac{1}{7}-\frac{1}{14}=\frac{1}{14}\)
\(7.\left[\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right):2\right]\)
\(7.\left[\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right):2\right]\)
\(7.\left(\frac{1}{3}-\frac{1}{13}\right):2\)
\(7.\frac{10}{39}:2=\frac{35}{39}\)
\(\frac{7}{15}+\frac{7}{35}+\frac{7}{63}+\frac{7}{99}+\frac{7}{143}\)
\(=\frac{7}{2}\cdot\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{7}{2}\cdot\left(\frac{1}{3}-\frac{1}{13}\right)\)
\(=\frac{7}{2}\cdot\frac{10}{39}\)
\(=\frac{35}{39}\)
(556/31-31/8)-(90/31-4)
3567/248-(90/31-4)
3567/248+34/31
3839/248
b1: a, 612.(15+19-34)=612.0=0
b,414.(37.4+23.4-240)=414.0=0
c,(517.125-518.25)+63:23=(517.53-518.52)+33=0+27=27
b2:a,143+7.(n-17)=206
===> 7.(n-17)=206-143=63
====>n-17=63:7=9
=====>n=9+17=26
vậy n=26
b,128-28:(15-n)=124
====>28:(15-n)=128-124=4
=====> 15-n=28:4=7
=====> n=15-7=8
vậy n=8
c,3n.2+48=210
====>3n.2=210-48=162
====>3n=162:2=81=34
====>n=4
vậy n=4
=(-156)+28+143
=(-156)+171
=15