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DQ
1
21 tháng 7 2015
9/10-1/90-1/72-1/56-1/42-1/30-1/20-1/12-1/6-1/2
=9/10-(1/9*10+1/8*9+...+1/1*2)
=9/10-(1/9-1/10+...+1-1/2)
=9/10-(-1/10+1)=9/10-9/10=0
BT
1
12 tháng 2 2020
= 9/1.10 + 1/9.10 + 1/8.9 + 1/7.8 + 1/6.7 +1/5.6 + 1/4.5 +1/3.4 +1/2.3 + 1/1.2
= 1/1.2 + 1/2.3 + 1/3.4 + ... + 9/1.10 ( viết ngược lại)
= 1-1/2 + 1/2 -1/3 + 1/3 +....-1/10
= 1 - 1/10
= 9/10
NA
2
6 tháng 8 2016
\(\frac{9}{10}-\frac{1}{90}-\frac{1}{72}-...-\frac{1}{6}-\frac{1}{2}\)
\(=\frac{9}{10}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}+\frac{1}{90}\right)\)
\(=\frac{9}{10}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{9}{10}-\left(1-\frac{1}{10}\right)\)
\(=\frac{9}{10}-\frac{9}{10}=0\)
2 tháng 8 2023
=9/10-(1/2+1/6+...+1/90)
=9/10-(1-1/2+1/2-1/3+...+1/9-1/10)
=9/10-9/10=0
H9
HT.Phong (9A5)
CTVHS
3 tháng 9 2023
a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-...+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\left(1-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\dfrac{9}{10}\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=90-89\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=1\)
\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}\)
\(\Rightarrow x=\dfrac{11}{25}\)
b) \(x\cdot9,85+x\cdot0,15=0,1\)
\(\Rightarrow x\cdot\left(9,85+0,15\right)=0,1\)
\(\Rightarrow x\cdot10=0,1\)
\(\Rightarrow x=\dfrac{0,1}{10}\)
\(\Rightarrow x=0,01\)
c) \(\dfrac{2}{5}+2022x=\dfrac{4}{10}\)
\(\Rightarrow\dfrac{2}{5}+2022x=\dfrac{2}{5}\)
\(\Rightarrow2022x=\dfrac{2}{5}-\dfrac{2}{5}\)
\(\Rightarrow2022x=0\)
\(\Rightarrow x=\dfrac{0}{2022}\)
\(\Rightarrow x=0\)
ND
Nguyễn Đức Trí
VIP
3 tháng 9 2023
a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\left(1\right)\)
Ta có :
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(\left(1\right)\Rightarrow\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=89\)
\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=90-89\)
\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)
\(\Rightarrow x+\dfrac{206}{100}=\dfrac{5}{2}:\dfrac{1}{2}\)
\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}.\dfrac{2}{1}\)
\(\Rightarrow x+\dfrac{103}{50}=5\)
\(\Rightarrow x=5-\dfrac{103}{50}\)
\(\Rightarrow x=\dfrac{250}{50}-\dfrac{103}{50}\)
\(\Rightarrow x=\dfrac{147}{50}\)
= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... 1/9.10
= 1 - 1/10
= 9/10
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