z đâu ra???

\(a.x^2-2xy+6y^2-12x+2y+41\)

\(=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5\)

\(=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y-1\right)^2\)

\(=\left(x-y-6\right)^2+5\left(y-1\right)^2\) ≥ \(0\)

\(b.\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}-\dfrac{2x}{y}-\dfrac{2y}{x}+3\)

\(=\dfrac{x^2}{y^2}-2.\dfrac{x}{y}+1+\dfrac{y^2}{x^2}-2.\dfrac{y}{x}+1+1\)

\(=\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2+1>0\)

Ta có:

\(VT=\frac{x}{y}+1+\frac{y}{x}+1-2\ge2\sqrt{\frac{x}{y}}+2\sqrt{\frac{y}{x}}-2\ge\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}+2\sqrt{\sqrt{\frac{x}{y}}.\sqrt{\frac{y}{x}}}-2=VP\)

Dấu "=" xảy ra khi \(x=y\)

Mk trả lời cho bn rùi đó

\(x^2-4xy+5y^2+2x-8y+5=\left(x-2y+1\right)^2+\left(y-2\right)^2\ge0\forall x,y\).

x² - 4xy + 5y² + 2x - 8y + 5

= x² + 4y² + 1 - 4xy + 2x  - 4y + y² - 2y + 1

= (x - 2y + 1)² + (y - 1)² ≥ 0

Theo bđt AM-GM 

\(x+\dfrac{1}{x}\ge2\sqrt{\dfrac{x.1}{x}}=2\Rightarrow\left(x+\dfrac{1}{x}\right)^2\ge4\)

\(y+\dfrac{1}{y}\ge2\sqrt{\dfrac{y.1}{y}}=2\Rightarrow\left(y+\dfrac{1}{y}\right)^2\ge4\)

Cộng vế với vế ta có đpcm 

Dấu ''='' xảy ra khi x = y = 1 

\(\left(x^2-y^2\right)^2=\left(x-y\right)^2\left(x+y\right)^2\) \(\Rightarrow\left\{{}\begin{matrix}x;y>0\\x+y< 1\end{matrix}\right.\)=> dccm sai = > người ra đề sai họăc người chép đề sai ;