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\(A=\frac{1\times111+2\times110+3\times109+...+111\times1}{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+111\right)}\)
\(A=\frac{1\times111+2\times110+3\times109+...+111\times1}{\left(1+1+...+1\right)+\left(2+2+...+2\right)+...+111}\)(\(111\)số hạng \(1\), \(110\)số hạng \(2\),...)
\(A=\frac{1\times111+2\times110+3\times109+...+111\times1}{1\times111+2\times110+3\times109+...+111\times1}\)
\(A=1\)
Đặt A=1/1*2+1/2*3+...+1/13*14
=> A=1-1/2+1/2-1/3+...+1/13-1/14
=> A=1-1/14
=> A=13/14
a) \(4\frac{3}{4}-1\frac{1}{2}\times1\frac{1}{3}=\frac{19}{4}-\frac{3}{2}\times\frac{4}{3}=\frac{19}{4}-2=\frac{11}{4}\)
b) \(3\frac{1}{2}-2\frac{2}{3}\div1\frac{5}{6}=\frac{7}{2}-\frac{8}{3}\div\frac{11}{6}=\frac{7}{2}-\frac{16}{11}=\frac{45}{22}\)
Giải
M = 1/1+ 2 + 1/1 + 2+ 3 + 1/1 + 2+ 3+ 4 + 1/1 + 2+ 3+ 4+ 5 =
M= 1/3 + 1/6+1/10+1/15
M = 10/30 + 5/30 + 3/30 + 2/30
M= 20/30 =2/3
Làm như vậy đúng ko
Ta dùng công thức \(1+2+...+n=\dfrac{n\times\left(n+1\right)}{2}\). Khi đó
\(\dfrac{1}{1+2}=\dfrac{1}{\dfrac{2\times3}{2}}=\dfrac{2}{2\times3}\);
\(\dfrac{1}{1+2+3}=\dfrac{1}{\dfrac{3\times4}{2}}=\dfrac{2}{3\times4}\);
\(\dfrac{1}{1+2+3+4}=\dfrac{1}{\dfrac{4\times5}{2}}=\dfrac{2}{4\times5}\);
...;
\(\dfrac{1}{1+2+3+...+2020}=\dfrac{1}{\dfrac{2020\times2021}{2}}=\dfrac{2}{2020\times2021}\).
\(\Rightarrow\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+\dfrac{1}{1+2+3+4}+...+\dfrac{1}{1+2+3+...+2020}\)
\(=\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+\dfrac{2}{4\times5}+...+\dfrac{2}{2020\times2021}\)
\(=2\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+...+\dfrac{1}{2020\times2021}\right)\)
\(=2\left(\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+...+\dfrac{2021-2020}{2020\times2021}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2021}\right)\)
\(=\dfrac{2019}{2021}\)
\(M=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}\)
\(M=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+\frac{1}{\left(1+5\right).5:2}\)
\(M=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}\)
\(M=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(M=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(M=2.\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(M=2.\frac{1}{2}-2.\frac{1}{6}\)
\(M=1-\frac{1}{3}=\frac{2}{3}\)
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}\)
\(A=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+\frac{1}{\left(1+5\right).5:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(A=2.\frac{1}{3}=\frac{2}{3}\)