Question

[khuathuuthien]

Giải phương trình nghiệm nguyên: 9x² + 3y² + 6xy - 6x +2y -35 =0

Answer 1

[Kiệt Nguyễn]

\(9x^2+3y^2+6xy-6x+2y-35=0\)

\(\Leftrightarrow\left(9x^2+6xy+y^2\right)-2\left(3x+y\right)+1+2y^2+4y+2=38\)

\(\Leftrightarrow\left(3x+y-1\right)^2+2\left(y+1\right)^2=38\)(*)

\(\Rightarrow\left(3x+y-1\right)^2=38-2\left(y+1\right)^2\le38\)

\(\Rightarrow-\sqrt{38}\le3x+y-1\le\sqrt{38}\)

Từ (*) suy ra 3x + y - 1 chẵn mà 3x + y - 1 nguyên nên \(3x+y-1\in\left\{\pm6;\pm4;\pm2;0\right\}\)

* Nếu \(3x+y-1=\pm6\)thì \(2\left(y+1\right)^2=2\Rightarrow y+1=\pm1\Rightarrow\orbr{\begin{cases}y=-2\\y=0\end{cases}}\)

Th1: \(3x+y-1=6\)

+) \(y=-2\Rightarrow x=3\)

+) \(y=0\Rightarrow x=\frac{7}{3}\left(L\right)\)

Th2: \(3x+y-1=-6\)

+) \(y=-2\Rightarrow x=-1\)

+) \(y=0\Rightarrow x=\frac{-5}{3}\left(L\right)\)

* Nếu \(3x+y-1=\pm4\)thì \(2\left(y+1\right)^2=22\left(L\right)\)

* Nếu \(3x+y-1=\pm2\)thì \(2\left(y+1\right)^2=34\left(L\right)\)

* Nếu 3x + y - 1 = 0 thì \(2\left(y+1\right)^2=38\left(L\right)\)

Vậy phương trình có 2 cặp nghiệm nguyên \(\left(x,y\right)\in\left\{\left(3;-2\right);\left(-1;-2\right)\right\}\)