=>x^2-4x+4+y^2-6y+9=0

=>(x-2)^2+(y-3)^2=0

=>x=2 và y=3

â, đánh giá về trái ta có

\(\sqrt{x^2-4x+5}=\sqrt{\left(x-2\right)^2+1}>=1\)

\(\sqrt{9y^2-6y+1}>=0\)

do đó dấu bằng xảy ra khi x=2 va y=1/3

phần b làm tương tự

b, VT <=2-1=1

\(x^2-4x+y^2-6y+15=2\)

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2-9y+9\right)+2=2\)

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

Vì \(\left(x-2\right)^2\ge0;\left(y-3\right)^2\ge0\) 

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

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(y-3\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

Vậy (x;y) = (2;3)

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2-6y+9\right)=0\)

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

Do \(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-3\right)^2\ge0\end{matrix}\right.\) ;\(\forall x;y\Rightarrow\left(x-2\right)^2+\left(y-3\right)^2\ge0\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x-2=0\\y-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

\(x^2-4x+y^2-6x+15=2\)

\(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2-6x+9\right)-4-9+15-2=0\)

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

Lại có :

\(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(y-3\right)^2\ge0\end{matrix}\right.\) \(\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow x=2;y=3\)

\(4x^2+4x+y^2-6y=24\)

\(\Leftrightarrow\left(4x^2+4x+1\right)+\left(y^2-6y+9\right)=34\)

\(\Leftrightarrow\left(2x+1\right)^2+\left(y-3\right)^2=34=3^2+5^2\)

\(TH1:\hept{\begin{cases}\left(2x+1\right)^2=3^2\\\left(y-3\right)^2=5^2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=8\end{cases}}\)

\(TH2:\hept{\begin{cases}\left(2x+1\right)^2=5^2\\\left(y-3\right)^2=3^2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\end{cases}}\)

Vay.....

\(4x^2+4x+y^2-6y=24\)

\(\Leftrightarrow4x^2+4x+y^2-6y-24=0\)

\(\Leftrightarrow\left(4x^2+4x+1\right)+\left(y^2-6y+9\right)-34=0\)

\(\Leftrightarrow\left(2x+1\right)^2+\left(y-3\right)^2=34\)

Mà \(34=3^2+5^2=\left(-3\right)^2+\left(-5\right)^2\)

Vì là nghiệm nguyên dương nên:

\(\left(2x+1\right)^2+\left(y-3\right)^2=3^2+5^2\)\(\Rightarrow\hept{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\orbr{\begin{cases}\\\end{cases}}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x+1=3\\y-3=5\end{cases}}\)hoặc     \(\orbr{\begin{cases}2x+1=5\\y-3=3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=2\\y=8\end{cases}}\)         hoặc     \(\orbr{\begin{cases}2x=4\\y=6\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\y=8\end{cases}}\)           hoặc      \(\orbr{\begin{cases}x=2\\y=6\end{cases}}\)

Vậy các cặp số (x;y) là: (1;8);(2;6)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(y^2-4y+4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)