x = 2017 nha

Ta có : x(x-2017)-2018x+2017.2018=0

=>x(x-2017)-2018(x-2017)=0

=>(x-2017)(x-2018)=0

=>x=2017;2018.

`x/(1.2)+x/(2.3)+x/(3.4)+.....+x/(2017.2018)=1`

`-> x/1 - x/2 +x/2-x/3+x/3-x/4+........+x/2017-x/2018=1`

`-> x-x/2018=1`

`-> 2017/2018 .x=1`

`-> x=2018/2017`

\(x^2+x-2017.2018\)
\(=x^2-2017x+2018x-2017.2018\)
\(=x\left(x+2018\right)-2017\left(x+2018\right)\)
\(=\left(x+2018\right)\left(x-2017\right)\)

đề bài sai hả

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

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

 x.2017.2016:2=2017.2018

 1008x=2018

 x=1009/504

\(a,\Leftrightarrow2x^2-10x-2x^2-x=-11\\ \Leftrightarrow-11x=-11\Leftrightarrow x=1\\ b,\Leftrightarrow x\left(x^2-6x+9\right)=0\\ \Leftrightarrow x\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-2018\right)-2017\left(x-2018\right)=0\\ \Leftrightarrow\left(x-2017\right)\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=2018\end{matrix}\right.\)