NV
1
K
Khách
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11 tháng 9 2017
\(A=\frac{3}{1}+\frac{3}{\frac{\left(2+1\right).2}{2}}+\frac{3}{\frac{\left(3+1\right).3}{2}}+....+\frac{3}{\frac{\left(100+1\right).100}{2}}\)
\(\Rightarrow A=\frac{3}{1}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+6.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(\Rightarrow A=\frac{3}{1}+\frac{6.99}{202}=\frac{297}{101}+\frac{3}{1}=\frac{600}{101}\)
kết quả k bik có sai k
I8
0
NH
2
3 tháng 6 2018
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+.....+\frac{3}{1+2+...+100}\)
\(=3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\)
\(=\frac{2}{2}.\left(3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\right)\)
\(=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)
\(=6.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)
\(=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=6.\left(1-\frac{1}{101}\right)\)
\(=6.\frac{100}{101}=\frac{600}{101}\)
Vậy \(A=\frac{600}{101}\)
QT
3 tháng 6 2018
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)
\(A=\frac{3.2}{2}+\frac{3.2}{\left(1+2\right).2}+\frac{3.2}{\left(1+2+3\right).2}+...+\frac{3.2}{\left(1+2+...+100\right).2}\)
\(A=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)
\(A=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(A=6\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(A=6\cdot\left(1-\frac{1}{101}\right)=6\cdot\frac{100}{101}=\frac{600}{101}\)
Vay A = ........
NN
1
12 tháng 5 2016
\(\frac{1}{2}:\frac{3}{7}+1\frac{1}{2}\cdot1\frac{2}{3}:x=1.\Rightarrow\frac{7}{6}+\frac{5}{2}:x=1\Rightarrow\frac{5}{2}:x=1-\frac{7}{6}=-\frac{1}{6}\Rightarrow x=-\frac{1}{6}:\frac{5}{2}=-\frac{1}{15}\)
LB
14 tháng 5 2017
\(A=5+\frac{5}{1+2}+\frac{5}{1+2+3}+...+\frac{5}{1+2+3+...+100}\)
N
14 tháng 5 2017
A = \(5+\frac{5}{1+2}+\frac{5}{1+2+3}+...+\frac{5}{1+2+3+..+100}\)
\(=5x\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(=5x\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{5050}\right)\)
\(=2x5x\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\right)\)
\(=10x\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{100x101}\right)\)
\(=10x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=10x\left(1-\frac{1}{101}\right)\)
\(=10x\frac{100}{101}\)
\(=\frac{1000}{101}\)
DP
16 tháng 6 2017
\(a.\) \(25,4-\frac{1}{4}\div x+1250\%=37,5\)
\(25,4-0.25\div x+12,5=37,5\)
\(25,4-0,25\div x=37,5-12,5\)
\(25,4-0,25\div x=25\)
\(0,25\div x=25,4-25\)
\(0,25\div x=0,4\)
\(x=0,25\div0,4\)
\(x=0,625\)
\(b.\)\(3\frac{1}{8}\div\left(\frac{2}{5}-x\right)\times\frac{8}{25}=3\)
\(\frac{3\times8+1}{8}\div\left(\frac{2}{5}-x\right)=3\div\frac{8}{25}\)
\(\frac{25}{8}\div\left(\frac{2}{5}-x\right)=\frac{75}{8}\)
\(\frac{2}{5}-x=\frac{25}{8}\div\frac{75}{8}\)
\(\frac{2}{5}-x=\frac{25}{8}\times\frac{8}{75}\)
\(\frac{2}{5}-x=\frac{1}{3}\)
\(x=\frac{2}{5}-\frac{1}{3}=\frac{6}{15}-\frac{5}{15}\)
\(x=\frac{1}{15}\)
16 tháng 6 2017
a)25,4-\(\frac{1}{4}\): x +1250%=37,5
25,4 - 0,25 : x + 12,5 = 37,5
25,4 - 0,25 : x =37,5 -12,5
25,4 - 0,25 : x = 25
0,25 : x = 25,4 -25
0,25 : x = 0,4
x = 0,25;0,4
x = 0,625
b) \(3\frac{1}{8}:\left(\frac{2}{5}-x\right)\times\frac{8}{25}=3\)
\(3,125:\left(0,4-x\right)\times0,32=3\)
\(3,125:\left(0,4-x\right)=3:0,32\)
\(3,125:\left(0,4-x\right)=9,375\)
\(0,4-x=3,125:9,375\)
\(0,4-x=\frac{1}{3}\)
\(x=0,4-\frac{1}{3}\)
\(x=\frac{1}{15}\)
D
12 tháng 5 2015
Đặt A = \(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+...+\frac{\frac{1}{2}}{1+2+3+....+100}\)
= \(\frac{1}{2}\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{100.101:2}\right)\)
= \(\frac{1}{2}\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\right)\)
= \(\frac{1}{2}.2\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)
= 1\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)
= \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)
\(A=\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+3+...+100}\)
\(\Rightarrow A=\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+\frac{3}{10}+...+\frac{3}{5050}\)
\(\Rightarrow A=\frac{2}{2}\left(\frac{3}{1}+\frac{3}{3}+\frac{3}{6}+\frac{3}{10}+...+\frac{3}{5050}\right)\)
\(\Rightarrow A=\frac{6}{2}+\frac{6}{6}+\frac{6}{10}+\frac{6}{20}+...+\frac{6}{10100}\)
\(\Rightarrow A=6\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{10}+\frac{1}{12}+...+\frac{1}{10100}\right)\)
\(\Rightarrow A=6\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\right)\)
\(\Rightarrow A=6\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A=6\left(1-\frac{1}{101}\right)\)
\(\Rightarrow A=6.\frac{100}{101}\)
\(\Rightarrow A=\frac{600}{101}\)