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\(x^2+2y^2+3xy+8=9x+10y\)
\(\Leftrightarrow4x^2+8y^2+12xy+32-36x-40y=0\)
\(\Leftrightarrow4x^2+12x\left(y-3\right)+\left(8y^2-40y+32\right)=0\)
\(\Leftrightarrow4x^2+12x\left(y-3\right)+9\left(y-3\right)^2-\left(y^2-14y+49\right)=0\)
\(\Leftrightarrow\left[2x-3\left(y-3\right)\right]^2-\left(y-7\right)^2=0\)
\(\Leftrightarrow\left[2x-3\left(y-3\right)-\left(y-7\right)\right].\left[2x-3\left(y-3\right)+\left(y-7\right)\right]=0\)
\(\Leftrightarrow\left(2x-4y+16\right)\left(2x-2y+2\right)=0\)
\(\Leftrightarrow\left(x-2y+8\right)\left(x-y+1\right)=0\)
-TH1: \(x-2y+8=0\) \(\Leftrightarrow x=2y-8\) thay vào pt đề cho tìm được x, y.
Tương tự cho TH2
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\(x^2=\left(y+1\right)^2+12\)
\(\Leftrightarrow\left(x-y-1\right)\left(x+y+1\right)=12\)
Do \(x,y\in N\)* nên \(x-y-1;x+y+1\inƯ\left(12\right)\) và \(x+y+1\ge1+1+1=3\)
TH1: \(x+y+1=12\Rightarrow x-y-1=1\)
\(\Leftrightarrow x=\dfrac{13}{2};y=\dfrac{9}{2}\) (ktm)
TH2:\(x+y+1=6;x-y-1=2\)
\(\Leftrightarrow x=4;y=1\) (thỏa mãn)
TH3: \(x+y+1=4;x-y-1=3\)
\(\Leftrightarrow x=\dfrac{7}{2};y=-\dfrac{1}{2}\) (ktm)
TH4: \(x+y+1=3;x-y-1=4\) (ktm)
Vậy \(x=4;y=1\)
\(x^2=y^2+2y+13\)
\(\Leftrightarrow x^2=y^2+2y+1+12\)
\(\Leftrightarrow x^2=\left(y+1\right)^2+12\)
\(\Leftrightarrow x^2-\left(y+1\right)^2=12\)
\(\Leftrightarrow\left(x-y-1\right)\left(x+y+1\right)=12\)
Vi x;y nguyên dương
\(\Rightarrow\left(x-y-1\right);\left(x+y+1\right)\in B\left(12\right)=\left\{1;2;3;4;6;12\right\}\left(x-y-1< x+y+1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+1\in\left\{12;6;4\right\}\\x-y-1\in\left\{1;2;3\right\}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\in\left\{\dfrac{13}{2};4;\dfrac{7}{2}\right\}\\y\in\left\{\dfrac{9}{2};1;-\dfrac{1}{2}\right\}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\) (x;y nguyên dương)
Vậy \(\left(x;y\right)\in\left(4;1\right)\) thỏa mãn đề bài
\(x^2-\left(y^2+2y+1\right)=12\)
\(x^2-\left(y-1\right)^2=12\)
\(\left(x-y+1\right)\left(x+y-1\right)=12\)
Mà x,y thuộc N nên dùng ước số, bạn làm tiếp nhé