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\(\dfrac{y+z+t-2020x}{x}=\dfrac{z+t+x-2020y}{y}=\dfrac{t+x+y-2020z}{z}=\dfrac{x+y+z-2020t}{t}=\dfrac{-2017\left(x+y+z+t\right)}{x+y+z+t}=-2017\\ \Leftrightarrow\left\{{}\begin{matrix}y+z+t-2020x=-2017x\\z+t+x-2020y=-2017y\\t+x+y-2020z=-2017z\\x+y+z-2020t=-2017t\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y+z+t=2x\\x+y+z+t=2y\\x+y+z+t=2z\\x+y+z+t=2t\end{matrix}\right.\\ \Leftrightarrow x=y=z=t=\dfrac{x+y+z+t}{2}=1010\\ \Leftrightarrow A=1010\left(2019-2020+2021-2022\right)=1010\left(-2\right)=-2020\)
\(P=\frac{2019xz}{xyz+2019xz+2019z}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{2019xz}{2019+2019xz+2019z}+\frac{y}{y\left(xz+z+1\right)}+\frac{z}{xz+z+1}\)
\(\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}=1\)
a)
TH1: x+2 =2019x+2020
x-2019x=2020-2
x(1-2019)=2018
x. (-2018)=2018
x=2018:(-2018)
x=-1
TH2: x+2 = -(2019x+2020)
x+2 =-2019x -2020
x+2019x = -2020-2
2020x=-2022
x=-2022:2020= - 1011/1010
b: \(=1^{2020}\cdot\left(-1\right)^{2021}+4\cdot1^{2020}\cdot\left(-1\right)^{2021}-2\cdot1^{2020}\cdot\left(-1\right)^{2021}\)
\(=1\cdot\left(-1\right)+4\cdot1\cdot\left(-1\right)-2\cdot1\cdot\left(-1\right)\)
=-1-4+2
=-3
Ta có: x = 2018 \(\Rightarrow x+1=2019\).
\(f\left(x\right)=x^6-2019x^5+2019x^4-...-2019+1\)
\(=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-...-\left(x+1\right)x+1\)
\(=x^6-x^6-x^5+x^5+x^4-...-x^2-x+1\)
\(=-x-1=-2018-1=-2019\)
Ta có:
\(\frac{x}{2}=\frac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
\(\Rightarrow A=\frac{2019x+2020y}{2019x-2020y}=\frac{2019.2k+2020.3k}{2019.2k-2020.3k}=\frac{10098k}{-2022k}=\frac{10098}{-2022}=\frac{-1683}{337}\)
Ta có:
\(\frac{x}{2}=\frac{y}{3}.\)
Đặt \(\frac{x}{2}=\frac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)
Lại có: \(A=\frac{2019x+2020y}{2019x-2020y}.\)
+ Thay \(x=2k\) và \(y=3k\) vào A ta được:
\(A=\frac{2019.2k+2020.3k}{2019.2k-2020.3k}\)
\(\Rightarrow A=\frac{4038k+6060k}{4038k-6060k}\)
\(\Rightarrow A=\frac{k.\left(4038+6060\right)}{k.\left(4038-6060\right)}\)
\(\Rightarrow A=\frac{4038+6060}{4038-6060}\)
\(\Rightarrow A=\frac{10098}{-2022}\)
\(\Rightarrow A=\frac{-1683}{337}.\)
Vậy \(A=\frac{-1683}{337}.\)
Chúc bạn học tốt!