Question

:-? Giải các phương trình nghiệm nguyên sau:

a) xy + 3x – y = 38 b)2x + 1 = y( x² + x + 1) c)yx² + ( y- 2) x + y – 1 = 0

d)2x² + y² - 2xy + 2y -6x + 5 = 0 e)x² - 4xy + 5y² =16

Answer 1

\(x^2-4xy+5y^2=16\)

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+y^2=16\)

\(\Leftrightarrow\left(x-2y\right)^2+y^2=16=4^2+0^2=0^2+4^2\)

\(TH1:\left\{{}\begin{matrix}\left(x-2y\right)^2=4^2\\y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4;x=-4\\y=0\end{matrix}\right.\)

\(TH2:\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\y^2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-4\\y=-2\end{matrix}\right.\)

Answer 2

\(xy+3x-y=38\)

\(\Leftrightarrow\left(xy-y\right)+\left(3x-3\right)=35\)

\(\Leftrightarrow y\left(x-1\right)+3\left(x-1\right)=35\)

\(\Leftrightarrow\left(x-1\right)\left(y+3\right)=35\)

Làm nốt

Answer 3

\(2x^2+y^2-2xy+2y-6x+5=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+2x-2y+1+\left(x^2-4xy+4\right)=0\)

\(\Leftrightarrow\left[\left(x-y\right)+2\left(x-y\right)+1\right]+\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(x-y+1\right)^2+\left(x-2\right)^2=0\)

\(\Leftrightarrow x=2;y=3\)