a: 5x-2y=7
=>5x=2y+7
=>\(x=\dfrac{2y+7}{5}\)
Để x nguyên và y nguyên thì \(2y+7⋮5\)
mà 7 chia 5 dư 2
nên 2y chia 5 dư 3
=>2y=5k+3(k\(\in\)Z)
=>\(y=\dfrac{5k+3}{2}\left(k\in Z\right)\)
mà y nguyên
nên 5k+3 chia hết cho 2
=>k lẻ
b:
ĐKXĐ: x>=0
\(x^2+2x+4=3\sqrt{x^3+4x}\)
=>\(3\sqrt{x\left(x^2+4\right)}=\left(x^2+4\right)+2x\)
=>\(\left(x^2+4\right)-3\sqrt{x}\cdot\sqrt{x^2+4}+2x=0\)
=>\(\left(x^2+4\right)-\sqrt{x}\cdot\sqrt{x^2+4}-2\sqrt{x}\cdot\sqrt{x^2+4}+2x=0\)
=>\(\sqrt{x^2+4}\left(\sqrt{x^2+4}-\sqrt{x}\right)-2\sqrt{x}\left(\sqrt{x^2+4}-\sqrt{x}\right)=0\)
=>\(\left(\sqrt{x^2+4}-2\sqrt{x}\right)\left(\sqrt{x^2+4}-\sqrt{x}\right)=0\)
=>\(\left[{}\begin{matrix}\sqrt{x^2+4}=2\sqrt{x}\\\sqrt{x^2+4}=\sqrt{x}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+4=4x\\x^2+4=x\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x^2-x+4=0\\x^2-4x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\\\left(x-2\right)^2=0\end{matrix}\right.\)
=>x-2=0
=>x=2(nhận)
