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A = 3/1 + 3/1+2 + 3/1+2+3 + 3/1+2+3+4 + ...+3/1+2+..+100
A = 3/1 + 3/3 + 3/6 + 3/10 +..+3/5050
A = 2/2 .( 3/1 + 3/3 + 3/6 + 3/10 +...+ 3/5050)
A = 6/2 + 6/6 + 6/12 + 6/20 +..+6/10100)
A = 6 .(1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +.. +1/100.101)
A = 6. (1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+1/100 - 1/101)
A = 6 (1 - 1/101)
A = 6 . 100/101
A = 600/101
\(\frac{2}{3}+\frac{3}{4}=\frac{17}{12}\)
\(\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)
\(\frac{2}{5}x\frac{3}{5}=\frac{6}{25}\)
\(\frac{9}{4}>\frac{9}{5}\)
\(\frac{2}{3}+\frac{3}{4}=\frac{8}{12}+\frac{9}{12}=\frac{17}{12}\)
\(\frac{1}{2}-\frac{1}{4}=\frac{2}{4}-\frac{1}{4}=\frac{1}{4}\)
\(\frac{2}{5}\cdot\frac{3}{5}=\frac{6}{25}\)
So sánh \(\frac{9}{4}\)và \(\frac{9}{5}\)
Vì tử số của hai phân số bằng nhau nên ta chỉ xét mẫu số, nếu mẫu số nào lớn hơn thì phân số đó bé hơn.
Vậy \(\frac{9}{4}\)và\(\frac{9}{5}\)mà\(4< 5\)nên\(\frac{9}{4}>\frac{9}{5}\)
\(A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{19.20}{2}}\)
=> \(A=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{19.20}\)
=> \(\frac{A}{2}=\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
=> \(\frac{A}{2}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
=> \(\frac{A}{2}=\frac{1}{3}-\frac{1}{20}\)
=> \(\frac{A}{2}=\frac{20-3}{20.3}\)
=> \(\frac{A}{2}=\frac{17}{60}\)
=> \(A=\frac{17}{30}\)
VẬY \(A=\frac{17}{30}\)
Ta có :\(\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+19}\)
\(=\frac{1}{3\times4}\times2+\frac{1}{4\times5}\times2+...+\frac{1}{19\times20}\times2\)
\(=2\times\left(\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{19\times20}\right)=2\times\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2\times\left(\frac{1}{3}-\frac{1}{20}\right)=2\times\frac{17}{60}=\frac{17}{30}\)
a) ( 1/2-1/3-1/6).(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10
0.(1/2+2/3+3/4+...+2017/2018) + 3/4.x = 9/10
0+3/4.x = 9/10
3/4.x = 9/10
x = 9/10: 3/4
x = 6/5
b) x + ( 3/1.3+3/3.5+...+3/13.15) = 11/5
x + 3/2. ( 1-1/3 + 1/3 - 1/5 + ...+ 1/13 - 1/15) = 11/5
x + 3/2. ( 1-1/15) = 11/5
x + 3/2.14/15 = 11/5
x + 7/5 = 11/5
x = 11/5 - 7/5
x = 4/5
\(\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)
\(=3\left(\frac{1}{\frac{1\cdot2}{2}}+\frac{1}{\frac{2\cdot3}{2}}+\frac{1}{\frac{3\cdot4}{2}}+...+\frac{1}{\frac{100\cdot101}{2}}\right)\)
\(=3\left(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+...+\frac{2}{100\cdot101}\right)\)
\(=6\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{100\cdot101}\right)\)
\(=6\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=6\left(1-\frac{1}{101}\right)=6-\frac{6}{101}=\frac{606-6}{101}=\frac{600}{101}\)