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\(=\dfrac{2}{2}\).(\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))
=2.(\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+...+\(\dfrac{2}{x.\left(x+1\right)}\))
=2.(\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\))
=2.[(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\))+(\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\))+(\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\))+...+(\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\))
=2.[\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)]
2.[(\(\dfrac{1}{3}\)-\(\dfrac{1}{3}\))+(\(\dfrac{1}{4}\)-\(\dfrac{1}{4}\))+...+(\(\dfrac{1}{x}\)-\(\dfrac{1}{x}\))+(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))]
=2.[0+0+...+0+(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))]
=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))
=2.(\(\dfrac{1.x+1-1.2}{2.x+1}\))
=2.(\(\dfrac{x+1-2}{2x}\))=2.\(\dfrac{x-1}{2x}\)=\(\dfrac{2.\left(x-1\right)}{2x}\)=\(\dfrac{2x-2}{2x}\)
\(\dfrac{2x-2}{2x}\)=\(\dfrac{2014}{2016}\)\(\Rightarrow\)(2x-2).2016=2014.2x=4032x-4032=4028x
\(\Rightarrow\)4032x-4028x=4x=4032\(\Rightarrow\)x=4032:4=1008
Đặt A=\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x.\left(x+1\right)}\)
\(A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}\)
\(A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x.\left(x+1\right)}\)
Tìm x \(\in\) Z biết:
1) \(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ...
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ....
Sửa đề:
So sánh:
\(A=\dfrac{10^{2015}+1}{10^{2016}+1}\) và \(B=\dfrac{10^{2016}+1}{10^{2017}+1}\)
Giải:
Ta thấy: \(\left\{{}\begin{matrix}A=\dfrac{10^{2015}+1}{10^{2016}+1}< 1\\B=\dfrac{10^{2016}+1}{10^{2017}+1}< 1\end{matrix}\right.\)
\(\Rightarrow\) Áp dụng tính chất \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) ta có:
\(B=\dfrac{10^{2016}+1}{10^{2017}+1}< \dfrac{10^{2016}+1+9}{10^{2017}+1+9}=\dfrac{10^{2016}+10}{10^{2017}+10}\)
\(=\dfrac{10\left(10^{2015}+1\right)}{10\left(10^{2016}+1\right)}=\dfrac{10^{2015}+1}{10^{2016}+1}\)
\(\Rightarrow\dfrac{10^{2016}+1}{10^{2017}+1}< \dfrac{10^{2015}+1}{10^{2016}+1}\)
Vậy \(B< A\)
Hay \(A>B\)
A=2^2016+3^2017
A=2^(4.504)+3^(4.504+1)
A=2^(4.504)+3^(4.504)+3^1
A= (...6)+(...1)+(...3)
A= (...0)
vậy chữ số tận cùng cua A là 0
bạn ơi (...3) đọc là chữ số tận cùng của 3
mình chắc chắn 100% là đúng. bài nay bạn học toán nâng cao lớp 6 dạng tìm 1 chữ số tận cùng là biết.
= (x^2017.x^-2017).(x^2016.x^-2016). ..... . (x^1.x^-1).x^0
= 1.1. .... .1.1 = 1
k mk nha
B = \(\frac{2015+2016+2017}{2016+2017+2018}=\frac{2016.3}{2017.3}=\frac{2016}{2017}\left(1\right)\)
Mà A = \(\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}.\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)=> A > B.
Vậy A > B .
Bạn Dont look at me
Bạn nên làm theo bạn ấy
Bạn k đúng cho bạn ấy. Bởi vì bạn ấy làm đúng
Theo mk là vậy
x=-2016
bn có thể giải thích giúp mk đc k????