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

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

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

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

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

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

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

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

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

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

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

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

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

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

Ta có : M = x² + 6x - 1

=> M = x² + 6x + 9 - 10

=> M = (x + 3)² - 10 

Mà : (x + 3)² \(\ge0\forall x\)

Nên M = (x + 3)² - 10 \(\ge-10\forall x\)

Vậy M_min = -10 , dấu "=" sảy ra khi x = -3

\(M=x^2+6x-1=\left(x^2+6x+9\right)-10=\left(x+3\right)^2-10\ge-10\)

Vậy \(MinM=-10\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x=-3\)

\(N=10y-5y^2-3=-5\left(y^2-2y+1\right)+5-3=-5\left(y-1\right)^2+2\le2\)

Vậy \(MaxN=2\Leftrightarrow-5\left(y-1\right)^2=0\Leftrightarrow y=1\)

\(P=x^2-4x+y^2-8y+6=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)

\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

Vậy \(MinP=-14\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\\left(y-4\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}}\)

\(N=10y-5y^2-3=-\left(5y^2-10y+5-8\right)=-\left[5\left(y^2-2y+1\right)-8\right]=8-5\left(y-1\right)^2\le8\)

maxN=8  khi (y-1)²=0 <=> y-1=0 <=> y=1

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\(P=x^2-4x+y^2-8y+6=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14=\left(x-2\right)^2+\left(y-4\right)^2-14\ge14\)

minP=14 khi (x-2)²=(y-4)²=0 <=>x-2=y-4=0 <=>x=2;y=4

a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)

\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)

\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)

Vậy Max_Q=10 khi x=2, y=-2

b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)

\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)

Vậy Max_A=14 khi x=-3

+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)

\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)

\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)

Vậy Max_B=5 khi x=-1/2, y=1/3

c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)

\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)

Vậy Min_P=2 khi x=1, y=-3

A=\(-x^2+2xy-4y^2+2x+8y-8=-\left(x^2-2xy+y^2-2x+1+2y\right)-\left(3y^2-6y+3\right)-4=-4-\left(x-y-1\right)^2-3\left(y-1\right)^2\le-4\)

=>Max A=-4<=>(x-y-1)²=0 và (y-1)²=0<=>x=2 y=1

de thi tu lam lay di, sao chui ban ay

bđt BCS : với các số a,b,x,y tùy ý : \(\left|ax+by\right|\le\sqrt{\left(a^2+b^2\right)\left(x^2+y^2\right)}\). Dấu = xảy ra <=> ay = bx