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\(17.8+51.4=34.4+51.4=4\left(51+34\right)=4.84=336\) \(2.2.3.5.19=\left(2.5\right).\left(3.19\right).2=10.2.57=570.2=1140\) \(54.275+825.15+275=54.275+45.275+275=275\left(54+45+1\right)=100.275=27500\) \(\frac{167.198+98}{198.168-100}=\frac{167.198+98}{198.167+198-100}=\frac{167.198+98}{167.198+98}=1\)
\(\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{2020}=1-\frac{1}{2020}=\frac{2019}{2020}\)
a) 17 x 8 + 51 x 4
= 17 x 4 x 2 + 17 x 3 x 4
= 17 x 4 x ( 2 + 3 )
= 14 x 4 x 5
= 14 x 20
= 280
b) 2 x 2 x 3 x 5 x 19
= ( 2 x 5 ) x ( 3 x 19 ) x 2
= 10 x 57 x 2
= 570 x 2
= 1140
c) 54 x 275 + 825 x 15 + 275
= 54 x 275 + 275 x 3 x 15 + 275 x 1
= 54 x 275 + 275 x 45 + 275 x 1
= 275 x ( 54 + 45 + 1 )
= 275 x 100
= 27500
d) 100 - 99 + 98 - 97 + 96 - 95 + 94 - 93 + ... + 4 - 3 + 2
= (100 - 99) + (98 - 97) + (96 - 95) + (94 - 93) + ... + (4 - 3) + 2
= (1 + 1 + ... + 1) + 2
( 49 số 1 )
= 49 + 2
= 51
k) 1,5 + 2,5 + 3,5 + 4,5 + 5,5 + 6,5 + 7,5 + 8,5
= ( 1,5 + 8,5 ) + ( 2,5 + 7,5 ) + ( 3,5 + 6,5 ) + ( 4,5 + 5,5 )
= 10 + 10 + 10 + 10
= 40
a) \(3.\frac{5}{4}\)\(-\frac{3^2}{4}\)\(=\frac{3}{2}\)
b)\(\frac{-21}{10}\)\(+\frac{21}{10}\)\(-\frac{3}{4}\)\(-\frac{3}{4}\)\(=\left(\frac{-21}{10}+\frac{21}{10}\right)-\left(\frac{3}{4}+\frac{3}{4}\right)\)
\(=0-\frac{3}{2}\)\(=\frac{-3}{2}\)
c) \(\frac{3}{4}\)\(+\frac{9}{5}-\frac{3}{2}-1\)\(=\left(\frac{3}{4}-\frac{3}{2}\right)+\left(\frac{9}{5}-1\right)\)\(=\frac{-3}{4}\)\(+\frac{4}{5}\)\(=\frac{1}{20}\)
\(N=\frac{4}{3}a-\left(\frac{1}{4}b+\frac{13}{12}b\right)\)
\(N=\frac{4}{3}a-\frac{4}{3}b\)
\(N=\frac{4}{3}\left(a-b\right)\)
\(N=\frac{4}{3}.\frac{3}{8}\)
\(N=\frac{1}{2}\)
\(\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)
\(\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)
\(\frac{3}{x}=\frac{-1}{2}\)
\(\Rightarrow3.2=\left(-1\right).x\)
\(\Rightarrow6=\left(-1\right).x\)
\(\Rightarrow x=6:\left(-1\right)\)
\(\Rightarrow x=-6\)
\(\frac{x}{2}-\frac{2}{y}=\frac{1}{2}\)
\(\Rightarrow\frac{x}{2}-\frac{1}{2}=\frac{2}{y}\)
\(\Rightarrow\frac{x-1}{2}=\frac{2}{y}\)
\(\Rightarrow\hept{\begin{cases}x-1=2\\2=y\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}\)
\(b,\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)
\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)
\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{8}{6}\)
\(\Rightarrow\frac{3}{x}=\frac{-3}{6}\)
\(\Rightarrow x\cdot(-3)=18\Rightarrow x=-6\)
Ta có
\(2017-\left(\frac{1}{4}+\frac{2}{5}+\frac{3}{6}+\frac{4}{7}+...+\frac{2017}{2020}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{4}+\frac{2}{5}+...+\frac{2017}{2020}\right)\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{2}{5}\right)+...+\left(1-\frac{2017}{2020}\right)\)
\(=\frac{3}{4}+\frac{3}{5}+....+\frac{3}{2020}\)
\(=\frac{3.5}{4.5}+\frac{3.5}{5.5}+\frac{3.5}{6.5}+...+\frac{3.5}{2020.5}\)
\(=3.5\left(\frac{1}{4.5}+\frac{1}{5.5}+\frac{1}{6.5}+...+\frac{1}{2020.5}\right)\)
\(=15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)\)
Thế vào ta có
\(\frac{15.\left(\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+...+\frac{1}{10100}\right)}{\frac{1}{20}+\frac{1}{25}+...+\frac{1}{10100}}=15\)
Được cập nhật 41 giây trước (17:23)
Ta có :
2017−(14 +25 +36 +47 +...+20172020 )
=(1+1+...+1)−(14 +25 +...+20172020 )
=(1−14 )+(1−25 )+...+(1−20172020 )
=34 +35 +....+32020
=3.54.5 +3.55.5 +3.56.5 +...+3.52020.5
=3.5(14.5 +15.5 +16.5 +...+12020.5 )
=15.(1
\(\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+...+\frac{4}{16\cdot18}\)
\(=2\cdot\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{16\cdot18}\right)\)
\(=2\cdot\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{16}-\frac{1}{18}\right)\)
\(=2\cdot\left(\frac{1}{2}-\frac{1}{18}\right)\)
\(=2\cdot\frac{4}{9}\)
\(=\frac{8}{9}\)
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