1.Tìm x biết: a) [3x-4]=x
b) [5x-3]=2x+1
2. Tính:
\(2.2016\)
1+\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+....+2016}\)
K
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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
\(A=\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}}\)
\(A=\frac{2.2016}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{2016.2017:2}}\)
\(A=\frac{4032}{\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{2}{2016.2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{2016}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+2\left(\frac{1}{2}-\frac{1}{2017}\right)}\)
\(A=\frac{4032}{1+\frac{2015}{2017}}\)
\(A=2017\)
7 tháng 9 2016
Mẫu số của A \(=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}\)
\(=\frac{1}{\left(1+0\right).2:2}+\frac{1}{\left(2+1\right).2:2}+\frac{1}{\left(3+1\right).3:2}+\frac{1}{\left(4+1\right).4:2}+...+\frac{1}{\left(2016+1\right).2016:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2.\left(1-\frac{1}{2017}\right)\)
\(=2.\frac{2016}{2017}=2.2016:2017\)
\(A=\left(2.2016\right):\left(2.2016:2017\right)\)
\(A=2.2016:2:2016.2017\)
\(A=2017\)
MA
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 \(=1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2016}\)
\(=\frac{1}{\left(0+1\right).2:2}+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+2016\right).2016:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(=2.\left(1-\frac{1}{2017}\right)\)
\(=2.\frac{2016}{2017}=2.2016:2017\)
\(A=\left(2.2016\right):\left(2.2016:2017\right)\)
\(A=2.2016:2:2016.2017\)
\(A=2017\)
LB
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=
AH
Tính
- \(\frac{2.2016}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+..+2016}}\)
1
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 \(=\frac{2.2016}{\frac{2.2016}{2017}}=2.2016.\frac{2017}{2.2016}=2017\)
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