Câu 11:
\(7\left(x-2017\right)^2=23-y^2\)
=>\(7\left(x-2017\right)^2+y^2=23\)
Vì x,y là các số tự nhiên nên \(y^2\) là số chính phương
mà y^2<=23
nên \(y^2\in\left\{0;1;4;9;16\right\}\)
TH1: \(y^2=0\)
=>y=0
=>\(7\left(x-2017\right)^2=23\)
=>\(\left(x-2017\right)^2=\dfrac{23}{7}\)
=>\(x\in\varnothing\)
TH2: \(y^2=1\)
=>y=1
\(7\left(x-2017\right)^2+y^2=23\)
=>\(7\left(x-2017\right)^2=23-1=22\)
=>\(\left(x-2017\right)^2=\dfrac{22}{7}\)
mà \(x\in N\)
nên \(x\in\varnothing\)
TH3: y^2=4
=>y=2
\(7\left(x-2017\right)^2=23-y^2\)
=>\(7\left(x-2017\right)^2=23-2^2=19\)
=>\(\left(x-2017\right)^2=\dfrac{19}{7}\)
mà x là số tự nhiên
nên \(x\in\varnothing\)
TH4: \(y^2=9\)
=>y=3
\(7\left(x-2017\right)^2=23-y^2\)
=>\(7\left(x-2017\right)^2=23-9=14\)
=>\(\left(x-2017\right)^2=2\)
mà x nguyên
nên \(x\in\varnothing\)
TH5: y^2=16
=>y=4
\(7\left(x-2017\right)^2=23-y^2\)
=>\(7\left(x-2017\right)^2=23-16=7\)
=>\(\left(x-2017\right)^2=1\)
=>\(\left[{}\begin{matrix}x-2017=1\\x-2017=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2018\left(nhận\right)\\x=2016\left(nhận\right)\end{matrix}\right.\)
