3y=z

\(S=5x^2+z^2-4xz-24x+16z+2080\)

\(S=\left(x-2z+8\right)^2+4x^2-40x+2080-8^2\)

\(S=\left(x-2z+8\right)^2+4\left(x-5\right)^2+2080-8^2-4.5^2\)

S_min =\(2080-8^2-4.5^2\)

đề thi học kỳ của mình cũng có câu này

S min là 4,5 ²

^{giáng sinh an lành}

\(S=4x^2-12xy+9y^2+32x-48y+64+x^2-8x+16+2000\)

\(S=\left(2x-3y\right)^2+16\left(2x-3y\right)+64+\left(x^2+8x+16\right)+2000\)

\(S=\left(2x-3y+8\right)^{^2}+\left(x-4\right)^2+2000\ge2000\)

MinS = 2000 khi x = 4 và y = 16/3

\(S=5x^2+9y^2-12xy+24x-48y+2028\)

\(=\left(9y^2-12xy-48y\right)+5x^2+24x+2028\)

\(=\left[\left(3y\right)^2-2.3y.\left(2x+8\right)+\left(2x+8\right)^2\right]+5x^2+24x+2028-\left(2x+8\right)^2\)\(=\left(3y-2x-8\right)^2+5x^2+24x+2028-4x^2-32x-64\)\(=\left(3y-2x-8\right)^2+\left(x^2-8x+16\right)+1948\)

\(=\left(3y-2x-8\right)^2+\left(x-4\right)^2+1948\ge1948\forall x;y\)Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=4\\y=\dfrac{16}{3}\end{matrix}\right.\)

\(4x^2+9y^2+64-12xy-48y+32x+x^2-8x+16+2\)

\(=\left(2x-3y+8\right)^2+\left(x-4\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\)x=4 và y=\(\frac{16}{3}\)

Vậy MINP=2 <=> x=4;y=16/3

\(A=5x^2+9y^2-12xy+24x-48y+81\)

\(A=4x^2+x^2+9y^2-12xy+32x-48y-8x+16+1+64\)

\(A=(4x^2+9y^2+64-12xy+32x-48y)+\left(x^2-8x+16\right)+1\)

\(A=[\left(2x\right)^2+\left(3y\right)^2+\left(8\right)^2-2.2x.3y-2.3y.8+2.2x.8]+\left(x^2-8x+16\right)+1\)

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

\(Do\) \(\left(2x-3y+8\right)^2\ge0\) \(và\) \(\left(x-4\right)^2\ge0\)

\(\Rightarrow A_{min}=1\)

ta có

5x^2+9y^2-12xy+8-48y+24x+72=0

<=>x^2-8x+16 + 4x^2+9y^2-12xy-48y+32x+64=0

<=> (x-4)^2+(2x-3y+8)^2=0

do(x-4)^2 ;(2x-3y+8)^2 \(\ge0\)

nên \(\left\{\begin{matrix}\left(x-4\right)^2=0\\\left(2x-3y+8\right)^2=0\end{matrix}\right.\)

<=> x=4 ;y=5,(3) (loại)

Vậy ko tồn tại cặp nghiệm nguyên