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a) Ta có:\(8\left(x-2019\right)^2⋮8\Rightarrow25-y^2⋮8\)\(\left(1\right)\)
Mặt khác: \(8\left(x-2019\right)^2\ge0\Rightarrow25-y^2\ge0\)\(\left(2\right)\)
Từ\(\left(1\right),\left(2\right)\)ta có: \(y^2=1;9;25\)
Xét:\(y^2=1\Rightarrow8\left(x-2019\right)^2=24\Rightarrow\left(x-2019\right)^2=3\left(ktm\right)\)
\(y^2=9\Rightarrow8\left(x-2019\right)^2=16\Rightarrow\left(x-2019\right)^2=2\left(ktm\right)\)
\(y^2=25\Rightarrow8\left(x-2019\right)^2=0\Rightarrow\left(x-2019\right)^2=0\Rightarrow x-2019=0\Rightarrow x=2019\left(tm\right)\)
Vậy \(y=5;x=2019\)
\(y=-5;x=2019\)
Ta có: \(25-y^2=8\left(x-2009\right)^2\)
mà\(8\left(x-2009\right)^2\ge0\Rightarrow25-y^2\ge0\left(1\right)\)
\(8\left(x-2009\right)^2⋮8\Rightarrow25-y^2⋮8\left(2\right)\)
từ\(\left(1\right),\left(2\right)\Rightarrow y^2\in\left\{1;9;25\right\}\)
\(+,y^2=1\Rightarrow8\left(x-2009\right)^2=24\Rightarrow\left(x-2009\right)^2=3\left(ktm\right)\)
\(+,y^2=9\Rightarrow8\left(x-2009\right)^2=16\Rightarrow\left(x-2009\right)^2=2\left(ktm\right)\)
\(+,y^2=25\Rightarrow8\left(x-2009\right)^2=0\Rightarrow\left(x-2009\right)^2=0\Rightarrow x-2009=0\Rightarrow x=2009\)
Vậy\(x=2009;y=5\)hoặc\(-5\)
a,Tìm x,y thuộc Z biết : 25-y2=8.(x-2009)2
b,Tìm x,y thuộc N biết : (2008.x+3y+1).(2008x+2008x+y)=225
Ta có \(25-y^2\le25\)
\(\Rightarrow8\left(x-2009\right)\le25\)
\(\Rightarrow\left(x-2009\right)^2\le3\)
\(\Rightarrow\left[\begin{matrix}\left(x-2009\right)^2=0\\\left(x-2009\right)^2=1\end{matrix}\right.\)
\(\left[\begin{matrix}x-2009=0\\\left[\begin{matrix}x-2009=1\\x-2009=-1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=2009\\\left[\begin{matrix}x=2010\\x=2008\end{matrix}\right.\end{matrix}\right.\)
Vậy x=2008 hoặc x=2009 hoặc x=2010
\(Q=3x^2\left(x+y+2\right)-y^2\left(x+y+2\right)+2\left(x+y+2\right)+2016\)
\(=2016\)
:>>
x+y=-2
Q=-6x2+2y2+6x2-2y2+2x+2y+2020
Q=2(x+y)+2020 = 2.(-2)+2020=2020-4=2016
Tick mk nhé