Gọi d là ước chung lớn nhất của x, y

\(\Rightarrow\left(x,y\right)=d\)

\(\Rightarrow x,y,z,t⋮d\)

\(\Rightarrow x=dx_1;y=dy_1;z=dz_1;t=dt_1;\)

Với \(x_1;y_1;z_1;t_1\in N;\left(x_1;y_1\right)=1\)

\(\Rightarrow14\left(x_1^2+y^2_1\right)=z_1^2+t_1^2⋮7\)

\(\Rightarrow z_1;t_1⋮7\)

\(\Rightarrow x_1^2+y_1^2⋮7\)

\(\Rightarrow x_1;y_1⋮7\)

Trái giả thuyết nên phương trình vô nghiệm nguyên.

Tham khảo

{x + y + z = 2
{2xy - z^2 = 4
<=> {z=2-y-x
       {z^2=2xy-4
<=>{z^2=4+y^2+x^2-4y+2xy-4x
      {z^2=2xy-4
=> 4+y^2+x^2-4y+2xy-4x=2xy-4
<=>8+y^2+x^2-4y-4x=0
<=> (x^2-4x+4)+(y^2-4y+4)=0
<=>(x-2)^2+(y-2)^2=0
<=>{(x-2)^2=0
      {(y-2)^2=0
<=>{ x=2
       {y=2
=>z=2-2-2=-2
vậy x=2,y=2,z=-2

\(\left\{{}\begin{matrix}\left(2m+1\right)x+y=2m-2\left(1\right)\\m^2x-y=m^2-3m\end{matrix}\right.\)

\(\Rightarrow\left(m^2+2m+1\right)x=m^2-m-2\)

\(\Rightarrow x=\dfrac{m^2-m-2}{m^2+2m+1}\left(m\ne-1\right)\)

\(\Rightarrow x=1+\dfrac{-3m-3}{m^2+2m+1}=1+\dfrac{-3\left(m+1\right)}{\left(m+1\right)^2}=1+\dfrac{-3}{m+1}\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow y=2m-2-\left(2m+1\right)\left(1-\dfrac{3}{m+1}\right)\)

\(\Rightarrow y=\dfrac{3m}{m+1}=3+\dfrac{-1}{m+1}\)

\(\Rightarrow x,y\in Z\left(m\in Z\right)\Leftrightarrow\left\{{}\begin{matrix}m+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\\m+1\inƯ\left(1\right)=\left\{\pm1\right\}\end{matrix}\right.\)

\(\Rightarrow m+1=\pm1\Leftrightarrow\left[{}\begin{matrix}m=0\left(tm\right)\\m=-2\left(tm\right)\end{matrix}\right.\)

a.

Với \(m=-2\Rightarrow\left\{{}\begin{matrix}x+y=2\\-2x+y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\3x=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=2-\dfrac{4}{3}=\dfrac{2}{3}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}x+y=2\\mx+y=m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\\left(m-1\right)x=m-2\end{matrix}\right.\)

Phương trình có nghiệm khi \(m\ne1\)

Khi đó: \(x=\dfrac{m-2}{m-1}=1-\dfrac{1}{m-1}\)

\(x\in Z\Rightarrow\dfrac{1}{m-1}\in Z\Rightarrow m-1=Ư\left(1\right)=\left\{-1;1\right\}\)

\(\Rightarrow m=\left\{0;2\right\}\)

Thay \(m=-2\) vào \(mx-y=m\) \(\Leftrightarrow-2x-y=-2\)

\(\left\{{}\begin{matrix}x+y=2\\-2x-y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2x-2y=-4\\-2x-y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2x-2y+2x+y=-4-\left(-2\right)\\x+y=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-y=-2\\x+y=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x+2=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=0\end{matrix}\right.\)

Vậy tập nghiệm có hệ pt : \(\left(x;y\right)=\left(0;2\right)\)