Đáp án cần chọn là: C

Ta có 2018(x−2018)=2018

x–2018=2018:2018

x–2018=1

x=2018+1

x=2019x

Vậy x=2019.

Đáp án là C

có: 2018.(x - 2018) = 2018

⇔ x - 2018 = 2018 : 2018

⇔ x - 2018 = 1

⇔ x = 2019

Đáp án là C

Ta có: 2018.(x - 2018) = 2018

⇔ x - 2018 = 2018 : 2018

⇔ x - 2018 = 1

⇔ x = 2019

a/ \(A=2018\cdot2018\)

\(=\left(2019-1\right)\cdot2018=2019\cdot2018-2018\)

\(B=2017\cdot2019\)

\(=\left(2018-1\right)\cdot2019=2018\cdot2019-2019\)

\(\Rightarrow A>B\)

b/ 

\(A=2018\cdot2019\)

\(=\left(2017+1\right)\cdot2019=2017\cdot2019+2019\)

\(B=2017\cdot2020\)

\(=2017\cdot\left(2019+1\right)=2017\cdot2019+2017\)

\(\Rightarrow A>B\)

Quên câu cuối ạ

c/ \(A=32\cdot53-31\)

\(=32\cdot53-32+1\)

\(B=53\cdot31-32\)

\(=53\cdot\left(32-1\right)-32=32\cdot53-32-53\)

có 1 > (-53)

\(\Rightarrow A>B\)

a) A > B

b) A > B

c) A > B

\(x+y+z=9\Leftrightarrow\left(x+y+z\right)^2=81\\ \Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=81\\ \Leftrightarrow xy+yz+xz=\dfrac{81-27}{2}=27\\ \Leftrightarrow x^2+y^2+z^2=xy+yz+xz\\ \Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2xz\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow x=y=z=\dfrac{9}{3}=3\left(x+y+z=9\right)\)

\(\Leftrightarrow\left(x-4\right)^{2018}+\left(y-4\right)^{2019}+\left(z-4\right)^{2020}\\ =\left(-1\right)^{2018}+\left(-1\right)^{2019}+\left(-1\right)^{2020}=1-1+1=1\)

Đề bài yêu cầu gì?