Tính A biết: A= 2^2016-(2^2015+2^2014+2^2013+.....+ 2^2+2)
Những câu hỏi liên quan
a) Tìm GTLN của biểu thức: A = -2x2 - 2xy - y2 + 2x - 2y + 20
b) Tính: \(\frac{2014.(2015^2+2016)-2016.(2015^2-2014)}{2014.(2013^2-2012)-2012.(2013^2+2014)}\)
b: \(=\dfrac{2014\cdot2015^2+2014\cdot2016-2016\cdot2015^2+2016\cdot2014}{2014\cdot2013^2-2014\cdot2012-2012\cdot2013^2-2012\cdot2014}\)
\(=\dfrac{2015^2\cdot\left(-2\right)+2\cdot\left(2015^2-1\right)}{2013^2\cdot\left(-2\right)-2\cdot\left(2013^2-1\right)}\)
\(=\dfrac{\left(-2\right)\cdot\left(2015^2-2015^2+1\right)}{\left(-2\right)\cdot\left(2013^2+2013^2-1\right)}=\dfrac{1}{2\cdot2013^2}\)
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14 tháng 2 2020 lúc 10:57
Tính các tổng sau:
a) A=1+(-2) + 3 +(-4) + ...+(- 2014) + 2015;
b) B= (-2) + 4 +(-6) + 8 ... +(-2014) + 2016;
c) 1+(-3) + 5 +(-7) + ... + 2013 +(-2015);
d) (-2015) + (-2014) + (-2013)+ ... + 2015 + 2016
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
\(B=\left(-2\right)+4+\left(-6\right)+8+\left(-10\right)+,...+\left(-2014\right)+2016\)
\(B=2+2+....+2\left(\text{504 số hạng 2}\right)=1008\)
c) 1 + ( -3 ) +5 + ( -7 ) + ...........+ 2013 + ( -2015 )
[ 1 + (-3 ) ] + [ 5 + -7 ] + .......... + [ 2013 + ( - 2015 ) ]
có số cặp là : [ ( 2015 - 1 ) : 2 + 1 ] : 2 = 504 ( cặp )
= -2 + -2 + -2 +..........+ -2
= -2 x 504
= -1008
Xem thêm câu trả lời
bài 2: Tính nhanh
a) A = 201^2
b) B= 498^2
c) C= 93. 107
d) D= 2016^2 - 2015. 2017
e) E= 2016^3 - 1
________
2016^2 + 2017
g) G= 2016^2 - 2015^2 + 2014^2 - 2013^2 + 2012^2 - 2011^2 + 2010^2 - 1^2
\(201^2=\left(200+1\right)^2=200^2+2.200.1+1^2=40000+400+1=40401\)
\(498^2=\left(500-2\right)^2=500^2-2.500.2+2^2=250000-2000+4=248004\)
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\(93.107=\left(100-7\right)\left(100+7\right)=100^2-7^2=10000-49=9951\)
\(2016^2-2015.2017=2016^2-\left(2016-1\right)\left(2016+1\right)=2016^2-2016^2+1^2=1\)
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Tính
A=3^2016 - 3^2015 + 3^2014 - 3^2013 + ......+ 3^2 - 3 + 1
B= 4^2016 - 4^2015 + 4^2014 - 4^2013 + ......+4^2 - 4 + 1
=>3A= 3^2017-3^2016+3^2015-...-3^2+3
=>3A+A=4A=3^2017+1=>A=\(\frac{3^{2017}+1}{4}\)
B tương tự nha
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Cho A= 1/2+1/3+1/4+..+1/2016
B= 2015/1+2014/2+2013/3+....+2/2014+1/2015. Tính B/A
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)
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Tính : A=1 /2016. 2015 +1/2015. 2014+1/2013. 2014+... +1/1. 2
Cho A = 1/2 + 1/3 + 1/4 + ... + 1/2017 B = 1/2015 + 2/2014 +3/2013 + ...+ 2015/2 + 2016/1 Tính B : A
Ta có: \(\dfrac{B}{A}=\dfrac{\dfrac{1}{2016}+\dfrac{2}{2015}+\dfrac{3}{2014}+...+\dfrac{2015}{2}+\dfrac{2016}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(1+\dfrac{2015}{2}\right)+\left(1+\dfrac{2014}{3}\right)+...+\left(1+\dfrac{2}{2015}\right)+\left(1+\dfrac{1}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
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Cho A = 1/2 + 1/3 + 1/4 + ...+ 1/2016
B = 1/2015 + 2/2014 + 3/2013 + ... + 2014/2 + 2015/1
Tính B : A
\(B=\left(\dfrac{1}{2015}+1\right)+\left(\dfrac{2}{2014}+1\right)+\left(\dfrac{3}{2013}+1\right)+...+\left(\dfrac{2014}{2}+1\right)+1\)
\(=\dfrac{2016}{2}+\dfrac{2016}{3}+...+\dfrac{2016}{2016}\)
=>B:A=2016
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