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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
Thay x = 2018 vào \(A=x^{2018}-2019x^{2017}+2019x^{2016}-2019x^{2015}+...+2019x^2-2019x-1\) ta được
\(2018^{2018}-2019.2018^{2017}+2019.2018^{2016}-2019.2018^{2015}+...+2019.2018^2-2019.2018-1\)
\(=\)\(2018^{2018}-2019\left(2018^{2017}-2018^{2016}+2018^{2015}-...-2018^2+2018\right)-1\)
Đặt \(B=2018^{2017}-2018^{2016}+2018^{2015}-...-2018^2+2018\)
\(2018B=2018^{2018}-2018^{2017}+2018^{2016}-...-2018^3+2018^2\)
\(2018B+B=\left(2018^{2018}-2018^{2017}+...+2018^2\right)+\left(2018^{2017}-2018^{2016}+...+2018\right)\)
\(2019B=2018^{2018}-2018\)
\(B=\frac{2018^{2018}-2018}{2019}\)
\(\Rightarrow\)\(A=2018^{2018}-2019.B-1\)
\(\Rightarrow\)\(A=2018^{2018}-\frac{2019\left(2018^{2018}-2018\right)}{2019}-1\)
\(\Rightarrow\)\(A=2018^{2018}-\left(2018^{2018}-2018\right)-1\)
\(\Rightarrow\)\(A=2018^{2018}-2018^{2018}+2018-1\)
\(\Rightarrow\)\(A=2018-1\)
\(\Rightarrow\)\(A=2017\)
Vậy giá trị của \(A=2017\) tại \(x=2018\)
Chúc bạn học tốt ~
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!