\(x^2-2x+5+y^2-4y=0\)

\(x^2-2\times x\times1+1^2-1^2+y^2-2\times y\times2+2^2-2^2+5=0\)

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

\(\left(x-1\right)^2\ge0\)

\(\left(y-2\right)^2\ge0\)

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

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

\(\Leftrightarrow x-1=y-2=0\)

\(\Leftrightarrow x=1;y=2\)

\(x^2+4y^2+13-6x-8y=0\)

\(\Leftrightarrow x^2-6x+9+4y^2-8y+4=0\)

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

Dấu = xảy ra khi

\(\orbr{\begin{cases}x-3=0\\2y-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\y=1\end{cases}}\)

1) x² - 2x + 5 + y² - 4y = 0

<=> x² - 2x + 1 + y² - 4y + 4 = 0

<=> ( x - 1 )² + ( y - 2 )² = 0

<=> \(\hept{\begin{cases}x-1=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)

2) x² + 4y² + 13 - 6x - 8y = 0

<=> x² - 6x + 9 + 4y² - 8y + 4 = 0

<=> ( x - 3 )² + ( 2y - 2 )² = 0

<=> \(\hept{\begin{cases}x-3=0\\2y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=1\end{cases}}\)

3) x² + y² + 6x - 10y + 34 = 0

<=> x² + 6x + 9 + y² - 10y + 25 = 0

<=> ( x + 3 )² + ( y - 5 )² = 0

<=> \(\hept{\begin{cases}x+3=0\\y-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=5\end{cases}}\)

Với điều kiện đã cho thì không tìm được $x,y,z$ cụ thể bạn nhé.

x2−6x+y2+10y+34=−(4z−1)2
x^2-6x+9+y^2+10y+25+(4z-1)^2=0x2−6x+9+y2+10y+25+(4z−1)2=0
(x-3)^2+(y+5)^2+(4z-1)^2=0(x−3)2+(y+5)2+(4z−1)2=0
{nghiempt}x-3=0\\y+5=0\\4z-1=0
{nghiempt}x=3\\y=-5\\z={1}{4}

x²-6x+y²+10y+34=-(4z-1)²

=>x²-6x+9+y²+10y+25+(4z-1)²=0=B

=>(x-3)²+(y+5)²+(4z-1)²=0

với mọi x,y,z ta có :

(x-3)²>=0

(y+5)²>=0

(4z-1)²>=0

=>(x-3)²+(y+5)²+(4z-1)²>=0

hay B>=0

dấu bằng xảy ra khi (x-3)²=0 => x-3=0  =>x=3

=>(y+5)²=0 =>y+5=0  =>y=-5

=>(4z-1)²=0 =>4z-1=0  => z=1/4

Vậy y=-5

\(x^2-6x+y^2+10y+34=-\left(4z-1\right)^2\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(y^2+10y+34\right)+\left(4z-1\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(y+5\right)^2+\left(4z-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(y+5\right)^2=0\\\left(4z-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-5\\z=\dfrac{1}{4}\end{matrix}\right.\)

Vậy........

Do đường tròn tiếp xúc với trục Ox nên R = d(I,Ox) = |yI|.

Phương trình trục Ox là y = 0

Đáp án D đúng vì: Tâm I(−3;\(\dfrac{-5}{2}\)) và bán kính R=\(\dfrac{5}{2}\). Ta có   

d(I, Ox) = |yI| = R.

\(4x^2+y^2-12x+10y+34=0\)

\(\Leftrightarrow4x^2-12x+9+y^2+10y+25=0\)

\(\Leftrightarrow\left(2x-3\right)^2+\left(y+5\right)^2=0\left(1\right)\)

mà \(\left\{{}\begin{matrix}\left(2x-3\right)^2\ge0,\forall x\\\left(y+5\right)^2\ge0,\forall y\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)^2=0\\\left(y+5\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-5\end{matrix}\right.\)

Ta có : \(4x^2+y^2-12x+10y+34=0\)

\(\Leftrightarrow4x^2-12x+9+y^2+10y+25=0\)

\(\Leftrightarrow\left(2x-3\right)^2+\left(y+5\right)^2=0\left(1\right)\)

Ta thấy : \(\left(2x-3\right)^2;\left(y+5\right)^2\ge0\)

Nên để (1) thoả mãn : 

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-5\end{matrix}\right.\)

Vậy........