20.18 x 2 và 2/5 + 201.8 x 97/22 x 220% + 0.006 x 2018
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\(20,18\times2\frac{2}{5}+201,8\times\frac{97}{22}\times220\%+0,006\times2018\)
\(=20,18\cdot\frac{12}{5}+20,18\cdot10\cdot\frac{97}{22}\cdot\frac{22}{10}+0,006\cdot100\cdot20,18\)
\(=20,18\cdot\frac{12}{5}+20,18\cdot\frac{970}{22}\cdot\frac{22}{10}+0,6\cdot20,18\)
\(=20,18\cdot\frac{12}{5}+20,18\cdot97+\frac{3}{5}\cdot20,18\)
\(=20,18\cdot\left(\frac{12}{5}+\frac{3}{5}\right)+20,18\cdot97\)
\(=20,18\cdot3+20,18\cdot97\)
\(=20,18\cdot\left(97+3\right)\)
\(=20,18\cdot100\)
\(=2018\)
20.18 x 89 + 20.18 x 0.2 + 20.18 x 0.5 + 20.18 x 0.25
= 20.18 ( 89 + 0.2 + 0.5 + 0.25 )
= 360 . 89,95
= 32382
Ủng hộ ae ơiii
20.18 x 89 + 20.18 x 0.2 + 20.18 x 0.5 + 20.18 x 0.25
= 20.18 x ( 89 + 0.2 + 0.5 + 0.25)
= 2018 x 89.95
= 1815.191
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
a/
\(a=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{17}\left(1+2+2^2+2^3\right)\)
Ta thấy
\(2\left(1+2+2^2+2^3\right)=2.15=30\)
\(\Rightarrow a=30+2^4.30+...+2^{16}.30⋮10\)
b/
Gọi tổng của 5 số TN liên tiếp là
n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10=5(n+2) chia hết cho 5
a) A= 1 - 3 + 5 - 7 + ... +97 - 99 +101 ( có 101 số )
A= ( 1 - 3 ) + ( 5 - 7 ) + ... + ( 97 - 99 ) + 101 ( có 50 nhóm )
A = - 2 + ( - 2 ) + .......... + ( - 2 ) + 101 ( có 50 số - 2 )
A = - 2 x 50 + 101
A = - 100 + 101
A = 1
A ) 1 - 3 + 5 - 7 +....+ 97 - 99 + 101
Dãy trên có số số hạng là :
( 101 - 1 ) : 2 + 1 = 51 ( số hạng )
Ta ghép mỗi bộ 2 số vậy có 25 bộ
Ta có :
1 - 3 + 5 - 7 +....+ 97 - 99 + 101
= ( 1 - 3 ) + ( 5 - 7 ) +....+ ( 97 - 99 ) + 101
= -2 + ( -2 ) + .....+ ( -2 ) + 101
Dãy trên có 25 số ( -2 )
Vậy tổng dãy trên là :
25 . ( -2 ) + 10 = -40
bn vt hỗn số kiểu j dzậy????
\(20,18\cdot2\frac{2}{5}+201,8\cdot\frac{97}{22}\cdot220\%+0,006\cdot2018\)
\(=2018\cdot\frac{1}{100}\cdot2\frac{2}{5}+2018\cdot\frac{1}{10}\cdot\frac{97}{22}+\frac{6}{1000}+2018\)
\(=2018\cdot\left(\frac{1}{100}\cdot2\frac{2}{5}+\frac{1}{10}\cdot\frac{97}{22}+6\cdot\frac{1}{1000}\right)\)
\(=2018\cdot\left[\frac{1}{1000}\cdot\left(6+100\cdot\frac{97}{22}+10\cdot2\frac{2}{5}\right)\right]\)
\(=\frac{1009}{500}\cdot\left[10\cdot\left(10\cdot\frac{97}{22}+\frac{12}{5}\right)\right]\)
\(=\frac{1009}{50}\cdot45\frac{12}{5}=\frac{239133}{250}\)