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\(A=\left(x^2+12x+36\right)+\left(y^2-2y+1\right)+3\\ A=\left(x+6\right)^2+\left(y-1\right)^2+3\ge3\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=1\end{matrix}\right.\)
\(A=x^2+y^2-2y+12x+40\)
\(=x^2+12x+36+y^2-2y+1+3\)
\(=\left(x+6\right)^2+\left(y-1\right)^2+3\ge3\forall x,y\)
Dấu '=' xảy ra khi x=-6 và y=1
\(B=\left(x-2y\right)^2+y^2+2x+6y+2046=\left[\left(x-2y\right)^2+2\left(x-2y\right)+1\right]+\left(y^2+10y+25\right)+2020=\left(x-2y+1\right)^2+\left(y+5\right)^2+2020\ge2020\)
\(minB=2020\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-11\\y=-5\end{matrix}\right.\)
C + A = B ⟹ C = B – A
C = (x2 + y – x2y2 – 1) – (x2 – 2y + xy + 1)
C = x2 + y – x2y2 – 1 – x2 + 2y – xy – 1
C = (x2– x2) + (y + 2y) – x2y2 – xy + ( - 1 – 1)
C = 0 + 3y – x2y2 – xy – 2
C = 3y – x2y2 – xy – 2
Ta có : A = x2 – 2y + xy + 1; B = x2 + y – x2y2 – 1
C = A + B = (x2 – 2y + xy + 1) + (x2 + y – x2y2 – 1)
C = x2 – 2y + xy + 1 + x2 + y – x2y2 – 1
C = (x2+ x2) + (– 2y + y) + xy – x2y2 + (1 – 1)
C = 2x2 – y + xy – x2y2 + 0
C = 2x2 – y + xy – x2y2
A – (xy + x2 – y2) = x2 + y2
A = (x2 + y2) + (xy + x2 – y2)
= x2 + y2 + xy + x2 – y2
= (x2 + x2) + (y2 – y2) + xy
= 2x2 + xy
\(A=x^2+y^2-8x-y+68=\left(x-4\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{207}{4}\ge\dfrac{207}{4}\)
\(minA=\dfrac{207}{4}\Leftrightarrow\)\(\left\{{}\begin{matrix}x=4\\y=\dfrac{1}{2}\end{matrix}\right.\)
\(A=x^2-8x+y^2-y+68\)
\(=x^2-8x+16+y^2-y+\dfrac{1}{4}+\dfrac{207}{4}\)
\(=\left(x-4\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{207}{4}\ge\dfrac{207}{4}\forall x,y\)
Dấu '=' xảy ra khi x=4 và \(y=\dfrac{1}{2}\)
x+2y = 1 => x = 1- 2y thay vào A là sẽ ra