Bạn nào ở tỉnh Bắc Giang cho mk xin vài tập đề toán 6 những năm 2012-2013, 2013-2014, 2014-2015, 2015-2016, 2016-2017
1) CMR : A=(n+2015)(n+2016) + n2 + n chia hết cho 2 với n ϵ N
2) So sánh :
P = \(\frac{2013}{2014^{2013}}+\frac{2014}{2015^{2014}}+\frac{2015}{2016^{2015}}+\frac{2016}{2017^{2016}}\) và
Q = \(\frac{2014}{2017^{2016}}+\frac{2013}{2016^{2015}}+\frac{2016}{2015^{2014}}+\frac{2015}{2014^{2013}}\)
A = (n + 2015)(n + 2016) + n2 + n
= (n + 2015)(n + 2015 + 1) + n(n + 1)
Tích 2 số tự nhiên liên tiếp luôn chia hết cho 2
=> (n + 2015)(n + 2015 + 1) chia hết cho 2
n(n + 1) chia hết cho 2
=> (n + 2015)(n + 2015 + 1) + n(n + 1) chia hết cho 2
=> A chia hết cho 2 với mọi n \(\in\) N (đpcm)
A = 2011 + 2012 + 2013 + 2014 + 2015 + 2016 + 2017 + 2018 + 2019
Số số hạng là:
\(\dfrac{2019-2011}{1}+1=8+1=9\left(số\right)\)
Tổng của dãy số là:
\(\left(2011+2019\right)\cdot\dfrac{9}{2}=18135\)
Cho A : \(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2014}\)
B :\(\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2017}\)
So sánh A và B
Chứng minh
\(Â=\dfrac{2013}{2013+2014}+\dfrac{2014}{2014+2015}+\dfrac{2015}{2015+2016}+\dfrac{2016}{2016+2017}< 2\)
\(\dfrac{2013}{2013+2014}< \dfrac{2013}{2013+2013}=\dfrac{1}{2}\)
Tương tự cộng theo vế suy ra đpcm
x-1/2012+x-2/2013 +x-3/2014=x-4/2015+x-5/2016+x-6/2017
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2013}+\dfrac{x-3}{2014}=\dfrac{x-4}{2015}+\dfrac{x-5}{2016}+\dfrac{x-6}{2017}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2012}+1\right)+\left(\dfrac{x-2}{2013}+1\right)+\left(\dfrac{x-3}{2014}+1\right)=\left(\dfrac{x-4}{2015}+1\right)+\left(\dfrac{x-5}{2016}+1\right)+\left(\dfrac{x-6}{2017}+1\right)\)
\(\Leftrightarrow\dfrac{x+2011}{2012}+\dfrac{x+2011}{2013}+\dfrac{x+2011}{2014}-\dfrac{x+2011}{2015}-\dfrac{x+2011}{2016}-\dfrac{x+2011}{2017}=0\)
\(\Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow x=-2011\)( do \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\))
tìm giá trị của biểu thức sau bằng cách hợp lí:
C= \(\dfrac{2014\left(2015^2+2016\right)-2016\left(2015^2-2014\right)}{2014\left(2013^2-2012\right)-2012\left(2013^2+2014\right)}\)
\(C=\dfrac{2014\left(2015^2+2016\right)-2016\left(2015^2-2014\right)}{2014\left(2013^2-2012\right)-2012\left(2013^2+2014\right)}\)
\(=\dfrac{2.2014.2016+2014.2015^2-2016.2015^2}{2014.2013^2-2012.2013^2-2.2012.2014}\)
\(=\dfrac{2.\left(2015+1\right)\left(2015-1\right)-2.2015^2}{2.2013^2-2.\left(2013+1\right)\left(2013-1\right)}\)
\(=\dfrac{2.\left(2015^2-1\right)-2.2015^2}{2.2013^2-2.\left(2013^2-1\right)}=\dfrac{-2}{2}=-1\)
1. tính:
1+2-3-4+5+6-7-8+9+10 -..............+2010 - 2011 - 2012 + 2013 + 2014 -2015 - 2016 + 2017
giúp mk nhé các bạn
thanks
mấy lần mk đăng mà chẳng bạn nào trả lời chán quá
đây nhé!!
1+2-3-4+5+6-7-8+9+10-........+2010-2011-2012+2013+2014-2015-2016+2017
=1+(2-3-4+5)+(6-7-8+9)+(10-11-12+13)+....+(2010-2011-2012+2013)+(2014-2015-2016+2017)
=1+0+0+0+.....+0+0
=1.
ĐÚNG THÌ CHO MINK NHA!!^_^
1+2-3-4+5+6-7-8+9+1-.........+2010-2011-2012+2013+2014-2015-2016+2017
tính
2011 + 2012 + 2013 + 2014 + 2015 + 2016 + 2017 + 2018 + 2019+ 2020 + 2021 +2022 +2023
2011+2012+2013+2014+2015+2016+2017+2018+2019+2020+2021+ 2022+2023 =(2011+2023)+(2013+2022)+...+(2016+2018)+2017 =4034+4034+4034+4034+4034+4034+2017 =4034x6+2017=26221
2011+2012+2013+2014+2015+2016+2017+2018+2019+2020+2021+2022+2023
=(2011+2023)+(2013+2022)+...+(2016+2018)+2017 =4034+4034+4034+4034+4034+4034+2017 =4034x6+2017=26221