Ta có: \(x^{2016}y^{2016}+5x^{2016}y^{2016}-3x^{2016}y^{2016}\)

Thay \(x=1;y=-1\) vào biểu thức

Ta được: \(1^{2016}.-1^{2016}+5.1^{2016}.-1^{2016}-3.2^{2016}.-1^{2016}\)

⇒ \(1.1.5.\left(-6\right)^{2016}\)

\(=5.\left(-6\right)^{2016}\)

Vậy giá trị của biểu thức \(x^{2016}y^{2016}+5x^{2016}y^{2016}-3x^{2016}y^{2016}\) tại \(x=1;y=-1\) là \(5.\left(-6\right)^{2016}\)

hay quá tớ cũng đang định hỏi

\(\frac{x}{20}=\frac{y}{9}=\frac{z}{6}\Leftrightarrow\frac{x}{20}=\frac{2y}{18}=\frac{4z}{24}=\frac{x-2y+4z}{20-18+24}=\frac{13}{26}=\frac{1}{2}\)

\(\Rightarrow x=\frac{1}{2}.20=10\)

\(\Rightarrow y=\frac{\frac{1}{2}.18}{2}=\frac{9}{2}\)

\(\Rightarrow z=\frac{\frac{1}{2}.24}{4}=3\)

ta có: \(\frac{x}{y}=\frac{z}{t}=\frac{z-2x}{2016y-2017t}=\frac{x-z}{y-t}=\frac{z-x}{2017\left(y-t\right)}\)

\(\Rightarrow2017\left(x-z\right)\left(y-t\right)=-\left(x-z\right)\left(y-t\right)\Rightarrow2017\left(y-t\right)=-\left(y-t\right)\)

\(\Rightarrow2018\left(y-t\right)=0\Rightarrow y=t\Rightarrow y^{2016}=t^{2016}\)

\(\Rightarrow y^{2016}-t^{2016}=0\)

x 20 = 14 20 + − 13 20 x 20 = 1 20       x   =   1

x = 1 or 0

\(^{x^{20}}\)=x

\(\Rightarrow\)\(^{x^{20}}\)-x =0

\(\Rightarrow\)x.(\(x^{19}\)-1)=0

\(\Rightarrow\)\(\hept{\begin{cases}x^{19}-1=0\\x=0\end{cases}}\Rightarrow\)\(\hept{\begin{cases}x=1\\x=0\end{cases}}\)

x20 là x²⁰ hay x.20 vậy bạn