Tính nhanh
\(2018\times2018-2019\times2017\)
Tính nhanh : \(\frac{2018\times2017-1}{2016\times2018+2017}\)
\(\frac{2018\times2017-1}{2016\times2018+2017}\)
\(=\frac{2018\times\left(2016+1\right)-1}{2016\times2018+2017}\)
\(=\frac{2018\times2016+2018-1}{2016\times2018+2017}\)
\(=\frac{2018\times2016+2017}{2016\times2018+2017}\)
\(=1\)
Tính nhanh:
a)\(\frac{131313}{151515}+ \frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)
b)\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{2017\times2018}\)
c)\(\frac{1}{3\times5}+\frac{1}{4\times7}+\frac{1}{7\times9}+...\frac{1}{2015\times2017}+\frac{1}{2017\times2018}\)
\(a,\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)
\(=\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(=13\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{7.9}\right)\)
\(=13\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(=13.\frac{2}{9}=\frac{26}{9}\)
\(b,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}=\frac{2017}{2018}\)
P/s :Dấu chấm là dấu nhân nha
\(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}\)
\(=\frac{13.10101}{15.10101}+\frac{13.10101}{35.10101}+\frac{13.10101}{63.10101}+\frac{13.10101}{99.10101}\)
\(=13.\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)
\(=\frac{13}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)
\(=\frac{13}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{13}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{13}{2}.\left(\frac{11}{33}-\frac{3}{33}\right)\)
\(=\frac{13}{2}.\frac{8}{33}\)
\(=\frac{52}{33}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=1-\frac{1}{2018}\)
\(=\frac{2017}{2018}\)
Sửa đề chút:
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{2015.2017}+\frac{1}{2017.2018}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2015}-\frac{1}{2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2017}\right)+\frac{1}{2017}-\frac{1}{2018}\)
B tự làm nốt nhé
tính bằng cách hợp lí
A = \(\dfrac{2018\times2017-1}{2016\times2018+2017}\)
B = \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)
$A=\dfrac{2018.2017-1}{2016.2018+2017}$
$=>A={2018.2016+2018-1}{2016.2018+2017}$
$=>A={2018.2016+2017}{2016.2018+2017}$
$=>A=1$
\(A=\dfrac{2018.2017-1}{2018.2016+2017}\)
\(A=\dfrac{2018.\left(2016+1\right)-1}{2018.2016+2017}\)
\(A=\dfrac{2018.2016+2018-1}{2018.2016+2017}\)
\(A=\dfrac{2018.2016+2017}{2018.2016+2017}=1\)
\(B=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}+\dfrac{1}{2187}\)
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^7}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\)
\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^6}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^7}\right)\)
\(\Rightarrow2B=1-\dfrac{1}{3^7}\Rightarrow B=\dfrac{1-\dfrac{1}{2187}}{2}=\dfrac{1093}{2187}\)
Chúc bạn học tốt!!!
Tính gía trị của biểu thức
\(\frac{\left(2015^2\times2025+31\times2016-1\right)\times\left(2015\times2020+4\right)}{2016^2\times2017\times2018\times2019\times2020}\)
tính
a, \(1\times2\times3\times...\times2018-1\times2\times3\times...\times2017^2\)
b,\(1500-\left\{5^2\times2^3-11\times\left[7^2-5\times2^3+8\times\left(11^2-121\right)\right]\right\}\)
\(\frac{2016\times2017+4034}{2018\times2019-4034}\) Tính
\(\frac{2016\times2017+4034}{2018\times2019-4034}=\frac{2016\times2017+2\times2017}{2018\times2019-2\times2017}\)
\(=\frac{\left(2016+2\right)\times2017}{2018\times2017+2\times2017-2\times2017}=\frac{\left(2016+2\right)\times2017}{2018\times2017+0}\)
\(=\frac{2018\times2017}{2018\times2017}=1\)
2020*2018+2019
__________________
2019*2019+2018
Tính nhanh
Trả lời :...............................................
\(\frac{4078379}{4078379}\)
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé Kim Râu La
so sán a và b biết a=\(\frac{2017\times2018-1}{2017\times2018}\)và b =\(\frac{2019\times2020-1}{2019\times2020}\)
\(Ta\)có :\(a\)=\(\frac{2017\cdot2018-1}{2017.2018}\)=\(\frac{2017.2018}{2017.2018}\)-\(\frac{1}{2017.2018}\)=1-\(\frac{1}{2017.2018}\)
\(b\)=\(\frac{2019.2020-1}{2019.2020}\)=\(\frac{2019.2020}{2019.2020}\)-\(\frac{1}{2019.2020}\)=1-\(\frac{1}{2019.2020}\)
Vì \(\frac{1}{2018.2019}\)> \(\frac{1}{2019.2020}\)nên \(a\)< \(b\)(sử dụng phần bù)
a) \(\frac{-3}{100}\)và\(\frac{2}{3}\) b)\(\frac{267}{-268}\)và\(\frac{-1347}{1343}\)c)\(\frac{2017\times2018-1}{2017\times2018}\)và\(\frac{2018\times2019-1}{2018\times2019}\)d)\(\frac{2017\times2018}{2017\times2018+1}\)và\(\frac{2018\times2019}{2018\times2019+1}\)
a) Ta có : \(\frac{-3}{100}< 0< \frac{2}{3}\)
\(\Rightarrow\frac{-3}{100}< \frac{2}{3}\)
b) Ta có : \(\frac{267}{268}< 1< \frac{1347}{1343}\)
\(\Rightarrow\frac{267}{268}< \frac{1347}{1343}\)
\(\Rightarrow\frac{267}{-268}< \frac{-1347}{1343}\)
c) Ta có : \(\frac{2017.2018-1}{2017.2018}=\frac{2017.2018}{2017.2018}-\frac{1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(\frac{2018.2019-1}{2018.2019}=\frac{2018.2019}{2018.2019}-\frac{1}{2018.2019}=1-\frac{1}{2018.2019}\)
mà \(2017.2018< 2018.2019\)
\(\Rightarrow\frac{1}{2017.2018}>\frac{1}{2018.2019}\)
\(\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)
\(\Rightarrow\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)
d) Ta có : \(\frac{2017.2018}{2017.2018+1}=\frac{2017.2018+1}{2017.2018+1}-\frac{1}{2017.2018+1}=1-\frac{1}{2017.2018+1}\)
\(\frac{2018.2019}{2018.2019+1}=\frac{2018.2019+1}{2018.2019+1}-\frac{1}{2018.2019+1}=1-\frac{1}{2018.2019+1}\)
mà \(2017.2018+1< 2018.2019+1\)
\(\Rightarrow\frac{1}{2017.2018+1}>\frac{1}{2018.2019+1}\)
\(\Rightarrow1-\frac{1}{2017.2018+1}< 1-\frac{1}{2018.2019+1}\)
\(\Rightarrow\frac{2017.2018}{2017.2018+1}< \frac{2018.2019}{2018.2019+1}\)