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BH
0
NT
1
HN
15 tháng 7 2019
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
\(\Leftrightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)
\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right)=1.\frac{1}{2}=\frac{1}{2}\)
\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}\)
\(\Leftrightarrow x+\frac{103}{50}=\frac{5}{2}.2\)
\(\Leftrightarrow x+\frac{103}{50}=5\)
\(\Leftrightarrow x=5-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{250}{50}-\frac{103}{50}\)
\(\Leftrightarrow x=\frac{147}{50}\)
NT
0
12 tháng 5 2019
A = 10.(1/1.2+5/2.3+...+89/9.10)
A/10 = 1/1.2+5/2.3+...+89/9.10
1.9 - A/10 = (1 - 1/1.2) + (1 - 5/2.3) +...+ (1 - 89/9.10)
9 - A/10 = 1/1.2 + 1/2.3 +....+ 1/9.10
9 - A/10 = 1 - 1/2 +1/2 -1/3 +...+ 1/9 -1 /10
9 - A/10 = 1 +0 +0+...... + 0 - 1/10
9 - A/10 = 1- 1/10
9 - A/10 = 9/10
A/10 = 9 - 9/10
A/10 = 81/10
A = (81/10) . 10
A = 81.
Vậy A = 81
LL
12 tháng 5 2019
Ta có : \(A=10\left(\frac{1}{1.2}+\frac{5}{2.3}+...+\frac{89}{9.10}\right)\)
\(\Rightarrow10\left(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}\right)\)
\(\Rightarrow10\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\right]\)
\(\Rightarrow10\left[9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\right]\)
\(\Rightarrow10\left[9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\right]\)
\(\Rightarrow10\left[9-\frac{9}{10}\right]\)
\(\Rightarrow10.\frac{81}{10}\)
\(\Rightarrow A=81\)
~\(Study\) \(well\)~
✰ᗪɾɑɕυɭɑ✰
11 tháng 5 2019
Đặt phép tính là A
Ta có :\(A=10.\left(\frac{1}{1.2}+\frac{5}{2.3}+...+\frac{89}{9.10}\right)\)
\(\Rightarrow A=10.\left(\frac{1}{2}+\frac{5}{6}+...+\frac{89}{90}\right)\)
\(\Rightarrow A=10.\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{90}\right)\right]\)
\(\Rightarrow A=10.\left[9-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\right)\right]\)
\(\Rightarrow A=10.\left[9-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\right]\)
\(\Rightarrow A=10.\left[9-\left(1-\frac{1}{10}\right)\right]\)
\(\Rightarrow A=10.\left[9-\frac{9}{10}\right]\)
\(\Rightarrow A=10.\frac{81}{10}\)
\(\Rightarrow A=81\)
~ Hok tốt ~
EJ
14 tháng 8 2017
a, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\) \(=-88\)
\(x+\dfrac{206}{100}=\dfrac{-5}{176}\)
\(x=\dfrac{-5}{176}-\dfrac{206}{100}\)
\(x=\dfrac{-9198}{4400}\)
JP
14 tháng 8 2017
a) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=90-89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)
\(x+\dfrac{206}{100}=5\)
\(x=5-\dfrac{206}{100}\)
\(x=\dfrac{147}{50}\)
Vậy \(x=\dfrac{147}{50}\)