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\(1-\left(11\frac{1}{2}-10,1+x\right):8\frac{2}{5}=0\)
=>\(1-\left(\frac{21}{2}-\frac{101}{10}+x\right):\frac{42}{5}=0\)
=> \(1-\left(\frac{2}{5}+x\right)=0\)
=>\(1=\frac{2}{5}+x\)
=>\(x=1-\frac{2}{5}\)
=>\(x=\frac{3}{5}\)
Vậy ..................

\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+....+\frac{2019}{2018.2019}\)
\(=\frac{2019}{1}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\left(1-\frac{1}{2019}\right)\)
\(=\frac{2019}{1}.\frac{2018}{2019}\)
\(=2018\)
\(A=\frac{2019}{2}+\frac{2019}{6}+\frac{2019}{12}+\frac{2019}{20}+\frac{2019}{30}+\frac{2019}{2018.2019}\)
\(A=\frac{2019}{1.2}+\frac{2019}{2.3}+\frac{2019}{3.4}+\frac{2019}{4.5}+\frac{2019}{5.6}+...+\frac{2019}{2018.2019}\)
\(A=2019.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\right)\)
\(A=2019.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\right)\)
\(A=2019.\left(1-\frac{1}{2019}\right)\)\(=2019.\frac{2018}{2019}=2018\)
Vậy A = 2018
-Dấu " . " là dấu nhân.

\(\frac{x}{5}=\frac{20}{x}\)
Ta có :
x . x = 20 . 5
x2 = 100
=> x = 10 hoặc x = -10

vì mỗi số hơn kém nhau 1.5 đơn vị nên
ta có công thức :
SSHxT1C:2=( (18.63-3.63):1.5+1)x(18.63+3.63):2=122.43
1002014nhân 20:100=200402,8
1002014 x 20 : 100=200402,8