\(10x^2+20y^2+24xy+8x-24y+51=\left(9x^2+24xy+16y^2\right)+\left(x^2+8x+16\right)+\left(4y^2-24y^2+36\right)-1=\left(3x+4y\right)^2+\left(x+4\right)^2+\left(2y-6\right)^2< 1\)

a: \(x^2+3y^2-4x+6y+7=0\)

\(\Leftrightarrow x^2-4x+4+3y^2+6y+3=0\)

\(\Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\)

\(\Leftrightarrow\left(x,y\right)=\left(-2;1\right)\)

\(x^2+3y^2-4x+6y+7=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(3y^2+6y+3\right)=0\\ \Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

\(3x^2+y^2+10x-2xy+26=0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(2x^2+10x+\dfrac{25}{8}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x^2+2\cdot\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x+\dfrac{5}{2}\right)^2+\dfrac{183}{8}=0\\ \Leftrightarrow x,y\in\varnothing\)

Sửa đề: \(3x^2+6y^2-12x-20y+40=0\)

\(\Leftrightarrow\left(3x^2-12x+12\right)+\left(6y^2-20y+\dfrac{50}{3}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y^2-2\cdot\dfrac{5}{3}y+\dfrac{25}{9}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y-\dfrac{5}{3}\right)^2+\dfrac{34}{3}=0\\ \Leftrightarrow x,y\in\varnothing\)

\(2\left(x^2+y^2\right)=\left(x+y\right)^2\\ \Leftrightarrow2x^2+2y^2=x^2+2xy+y^2\\ \Leftrightarrow x^2-2xy+y^2=0\\ \Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x-y=0\Leftrightarrow x=y\)

xy là x.y hay là x và y vậy bn

X và y là số nguyên phải ko

\(a,4x^2+9y^2+4x-24y+17=0\)

\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)

\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)

\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)