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\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2016}\)
\(A=\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{2016.2017:2}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2016.2017}\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{2017}\right)\)
\(A=2.\frac{2015}{4034}=\frac{2015}{2017}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
\(N=\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}\)
\(N=1+\left(\frac{1}{2016}+1\right)+\left(\frac{2}{2015}+1\right)+\left(\frac{3}{2014}+1\right)+...+\left(\frac{2015}{2}+1\right)\)
\(N=\frac{2017}{2017}+\frac{2017}{2016}+\frac{2017}{2015}+\frac{2017}{2014}+...+\frac{2017}{2}\)
\(N=2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{M}{N}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}{2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)}=\frac{1}{2017}\)
Ta có: \(B=\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}\)
\(B=1+\left(\frac{1}{2016}+1\right)+\left(\frac{2}{2015}+1\right)+\left(\frac{3}{2014}+1\right)+...+\left(\frac{2015}{2}+1\right)\)
\(B=\frac{2017}{2017}+\frac{2017}{2016}+\frac{2017}{2015}+\frac{2017}{2014}+...+\frac{2017}{2}\)
\(B=2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)}\)
\(\Rightarrow\frac{A}{B}=\frac{1}{2017}.\)
Chúc bạn học tốt!
Này Vũ Minh Tuấn, mk cũng có 1 bài cũng gần giống như thế này nhưng khác 1 tí cậu giải giúp mk vs
B = \(1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}\right)\)
Xét \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2016^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}=1-\frac{1}{2016}=\frac{2015}{2016}\)
Do đó B > 1 - \(\frac{2015}{2016}=\frac{1}{2016}\)
Bài 2:
a, 1/3 + 1/2 : x = -4
=> 1/2 : x = -4 - 1/3
=> 1/2 : x = -13/3
=> x = 1/2 ; -13/3
=> x = -3/26
Vậy x = -3 / 26
Bài 2:
b, x2 - 4x = 0
=> x.(x - 4) =0
=> x=0 hoặc x - 4 = 0
x - 4= 0 => x=4
Vậy x=0 và x=4