=> (x + x + x + .... + x) . (1 + 2 + 3 + ..... + 2016) = 2017.2018

=> 2016x . 1008 . 2017 = 2017 . 2018

=> 2016x . 1008 = 2018

=> /////////////

Cậu viết sai đề bài rồi!!!

=>(1+2+3+......+2016).x=4070306

=>2033136.x=4070306

=>x=4070306:2033136

=>x=\(\frac{1009}{504}\)

x+2x+3x+......+2016x=2017.2018

\(\Rightarrow\)x.(1+2+3+....+2016)=2017.2018

\(\Rightarrow\)x.2033136=2017.2018

\(\Rightarrow\)x=(2017.2018):2033136

x=\(\dfrac{1009}{504}\)

Chúc bạn học tốt!

Bài 2:

\(A=\dfrac{2016-x}{6-x}=\dfrac{2010+6-x}{6-x}=\dfrac{2010}{6-x}+\dfrac{6-x}{6-x}=1+\dfrac{2010}{6-x}\)

\(A\) đạt \(Max\) khi và chỉ khi \(6-x\) lớn nhất

*)Nếu \(x>6\Rightarrow6-x< 0\Rightarrow\dfrac{2010}{6-x}< 0\)

*)Nếu \(x< 6\Rightarrow6-x>0\Rightarrow\dfrac{2010}{6-x}>0\)

Nên \(\dfrac{2010}{6-x}\) lớn nhất khi \(6-x\) là số nguyên dương nhỏ nhất

\(\Rightarrow6-x=1\Rightarrow x=5\). Khi đó

\(A=1+\dfrac{2010}{6-5}=1+\dfrac{2010}{1}=1+2010=2011\)

Vậy \(A_{Max}=2011\) khi \(x=5\)

1/ Ta có: \(x+2x+3x+...=2016x=2017.2018\)

\(\Rightarrow2016x=4070306\)

\(\Rightarrow x=\dfrac{4070306}{2016}\)

Vậy \(x=\dfrac{4070306}{2016}\).

Câu 1:Sửa lại đề

x+2x+3x+...+2016x=2017.2018

x(1+2+3+.....+2016)=2017.2018

x.2033136=2017.2018

x=2017.2018:2033136

x=\(\dfrac{1009}{504}\)

Số số hạng là (2016-1):1+1=2016 số

Tổng là (2016+1)*2016/2=2033136

=>2033136x=2017*2018

=>1008x=2018

=>x=2018/1008

1) \(x+2\cdot x+3\cdot x+4\cdot x+...+2016\cdot x=2016\cdot2017\)

\(=x\cdot\left(2+3+4+...+2016\right)=4066272\)

\(=x\cdot2033136=4066272\)

\(\Rightarrow x=\frac{4066272}{2033136}=2\)

2) \(x-2\cdot x+3\cdot x-4\cdot x+...-2016\cdot x=2016\cdot2017\)

\(x\cdot\left\{\left(-2+\left(-4\right)+\left(-6\right)+...+\left(-2016\right)\right)+\left(1+3+5+7+...+2015\right)\right\}=4066272\)

\(x\cdot\left\{\left(-1017072\right)+1016064\right\}=4066272\)

\(x\cdot\left(-1008\right)=4066272\)

\(\Rightarrow x=-\frac{4066272}{1008}=-4034\)