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A = \(\left(2018-2017\right)\left(2018+2017\right)+\left(2016-2015\right)\left(2016+2015\right)+\)
\(...+\left(2-1\right)\left(2+1\right)\)
A = \(2018+2017+2016+2015+...+2+1\)
A = \(\dfrac{2018.2019}{2}=2037171\)
A=20182+20162+20142+...+42 +22-(20172 +20152+20132+...+ 32 + 1)
A=(2018²-2017²)+(20162-20152)+(2014²-2013²)+...+(2² −1²)
A=2018+2017+2016+2015+2014+2013+...+2+1
\(A=\dfrac{2018\left(2018+1\right)}{2}=\text{2 037 171}\)
\(B=2016^2+2017^2-2\\ B=2016^2-1+2017^2-1\\ B=\left(2016-1\right)\left(2016+1\right)+\left(2017-1\right)\left(2017+1\right)\\ B=2015.2017+2016.2018=A\)
ta có 2015 x 2017 >2017^2 -2
2016 x 2018 > 2016^2
=> A> B
\(2018^2-2017^2+2016^2-2015^2+...+2^2-1^2\)
\(=\left(2018+2017\right)\left(2018-2017\right)+\left(2016+2015\right)\left(2016-2015\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=4035+4031+...+3\)
Từ 3 đến 4035 có số lượng số hạng là:
\(\left(4035-3\right):4+1=1009\)
Ta có:
\(4035+4031+....+3\)
\(=\dfrac{\left(4035+3\right).1009}{2}=2037171\)
Chúc bạn học tốt!!!
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)
<=>\(\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{2017}-1+\dfrac{x+4}{2018}-1\)
<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)
<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)
<=>\(\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
vì 1/2015+1/2016-1/2017-1/2018 khác 0
=>x-2014=0<=>x=2014
vậy.....................
chúc bạn học totts ^^
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)
\(\Leftrightarrow\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{x017}-1+\dfrac{x+4}{2018}-1\)
\(\Leftrightarrow\dfrac{x+1-2015}{2015}+\dfrac{x+2-2016}{2016}=\dfrac{x+3-2017}{2017}+\dfrac{x+4-2018}{2018}\)\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)
\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
Vì: \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x=2014\)
Vậy........
20182 - 20172 + 20162 - 20152 + ... + 22 - 12
= (2018+2017)(2018-2017) + (2016+2015)(2016-2015) + ... + (2+1)(2-1)
= 2018 + 2017 + 2016 + 2015 + ... + 2 + 1
= \(\dfrac{\left(1+2018\right).2018}{2}=2037171\)