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8 tháng 9 2016

\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)

\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{2016.2017:2}}\)

\(A=\frac{4032}{1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{3.4}+...+\frac{2}{2016.2017}}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2016}-\frac{1}{2017}\right)}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}\)

\(A=\frac{4032}{1+2\left(\frac{2015}{2017}\right)}\)

\(\Rightarrow A=2017\)

8 tháng 9 2016

\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)

\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{2016.2017:2}}\)

\(A=\frac{4032}{\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{2}{2016.2017}\right)}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2016}-\frac{1}{2017}\right)}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}\)

\(A=\frac{4032}{1+\frac{2015}{2017}}\)

\(A=2017\)

30 tháng 9 2016

Ta có:1+2+3+...+n=\(\frac{n.\left(n+1\right)}{2}\)

=>B=\(\frac{2.2016}{\frac{2}{2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2016.2017}}\)

=>B=\(\frac{2016}{\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}}\)

=>B=\(\frac{2016}{\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+...+\left(\frac{1}{2016}-\frac{1}{2017}\right)}\)

=>B=

7 tháng 9 2016

Mẫu số của A \(=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}\)

\(=\frac{1}{\left(1+0\right).2:2}+\frac{1}{\left(2+1\right).2:2}+\frac{1}{\left(3+1\right).3:2}+\frac{1}{\left(4+1\right).4:2}+...+\frac{1}{\left(2016+1\right).2016:2}\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(=2.\left(1-\frac{1}{2017}\right)\)

\(=2.\frac{2016}{2017}=2.2016:2017\)

\(A=\left(2.2016\right):\left(2.2016:2017\right)\)

\(A=2.2016:2:2016.2017\)

\(A=2017\)

7 tháng 9 2016

\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}}\)

\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+..+\frac{1}{2016.2017:2}}\)

\(A=\frac{4032}{1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{2016.2017}\right)}\) .

\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)}\)

\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}=\frac{4032}{1+\frac{2015}{2017}}\)

\(A=2017\)

7 tháng 9 2016

Mẫu số \(=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}\)

\(=\frac{1}{\left(0+1\right).2:2}+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+2016\right).2016:2}\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(=2.\left(1-\frac{1}{2017}\right)\)

\(=2.\frac{2016}{2017}=2.2016:2017\)

\(A=\left(2.2016\right):\left(2.2016:2017\right)\)

\(A=2.2016:2:2016.2017\)

\(A=2017\)

19 tháng 10 2016

Mẫu số = \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2016}\)

\(=1+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2016\right).2016:2}\)

\(=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{1}{2016.2017}\)

\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)\)

\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(=2.\left(1-\frac{1}{2017}\right)\)

\(=\frac{2.2016}{2017}\)

Vậy phân số đề bài cho \(=\frac{2.2016}{\frac{2.2016}{2017}}=2.2016.\frac{2017}{2.2016}=2017\)

13 tháng 11 2016

mẫu:\(\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2016}\)

=\(1+\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+...+\frac{1}{\frac{\left(2016+1\right).2016}{2}}\)

=\(1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2016.2017}\)

=\(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{2016.2017}\right)\)

=\(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)

=\(1+2\left(\frac{1}{2}-\frac{1}{2017}\right)\)

=\(1+1-\frac{2}{2017}\)

=\(\frac{4032}{2017}\)

=>Biểu thức:\(\frac{4032}{\frac{4032}{2017}}\)

=\(2017\)

14 tháng 11 2016

Ta có công thức tổng quát với n tự nhiên là

\(1+2+...+n=\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow\frac{1}{1+2+...+n}=\frac{2}{n\left(n+1\right)}\)

Áp dụng công thức vào bài toán ta được

\(\frac{2.2016}{\frac{1}{1}+\frac{1}{1+2}+..+\frac{1}{1+2+...+2016}}=\frac{2.2016}{\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{2016.2017}}\)

\(=\frac{2.2016}{2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\right)}=\frac{2.2016}{2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\right)}\)

\(=\frac{2.2016}{2\left(1-\frac{1}{2017}\right)}=\frac{2.2016}{\frac{2.2016}{2017}}=2017\)

3 tháng 5 2018

1)

a)Xét 3x-4>=0 => x>=\(\frac{4}{3}\), ta có:

               3x-4 = x

               x=2( nhận )

Xét 3x-4<0  => x<\(\frac{4}{3}\)

             -3x+4 = x

              x = 1 ( nhận )

b) tương tự

6 tháng 9 2016

Xét mẫu số của F : 

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+..+2016}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{2016\cdot2017}{2}}\)

\(=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2016.2017}\right)=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)

\(=1+2\left(\frac{1}{2}-\frac{1}{2017}\right)=2-\frac{2}{2017}=\frac{4032}{2017}\)

Suy ra : \(F=\frac{2.2016}{\frac{4032}{2017}}=\frac{2.2016.2017}{4032}=2017\)