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        2x² + 3y² + 4x =19

 <=> 2(x+1)² + 3y² = 21

  => 3y²   =< 21  <=> y²  =<  7 => y= { -2;-1:0:1:2}

\(2x\left(3y-2\right)+\left(3y-2\right)=-55\)

\(\left(3y-2\right)\left(2x+1\right)=-55=1.\left(-55\right)=\left(-1\right).55=\left(-5\right).11=5.\left(-11\right)\)

Vậy \(\left(y,x\right)=\left\{\left(1;-28\right),\left(-1;5\right)\right\}\)

Câu hỏi của nganhd - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo nhé!

\(2x^2+3y^2+4x=19\)

\(2x^2+4x=19-3y^2\)

\(2\left(x+1\right)^2=3\left(7-y^2\right)\)

Vì \(2\left(x+1\right)^2⋮2\) nên \(3\left(7-y^2\right)⋮2\) hay \(7-y^2⋮2\Rightarrow y^2\) lẻ(1)

Ta có: \(\left(x+1\right)^2\ge0\Rightarrow7-y^2\ge0\Rightarrow y^2\le7\)\(\Rightarrow y^2\in\left\{1;4\right\}\)(2)

Từ (1) và (2), ta suy ra: \(y^2=1\)\(\Rightarrow y\in\left\{-1;1\right\}\)

Ta có: \(2\left(x+1\right)^2=3\left(7-y^2\right)\)

\(2\left(x+1\right)^2=18\)

\(\left(x+1\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x+1=3\Rightarrow x=2\\x+1=-3\Rightarrow x=-4\end{matrix}\right.\)

Các cặp số nguyên \(\left(x;y\right)=\left\{\left(2;1\right);\left(2;-1\right);\left(-4;1\right);\left(-4;-1\right)\right\}\)