\(A=\frac{x^2-2x+2018}{x^2}=1-\frac{2}{x}+\frac{2018}{x^2}\)

\(=2018t^2-2t+1\left(\frac{1}{x}=t\right)\)

\(=2018\left(t^2-\frac{2t}{2018}+\frac{1}{2018}\right)\)

\(=2018\left(t-\frac{1}{2018}\right)^2+\frac{2017}{2018}\ge\frac{2017}{2018}\)

B > = 0 

Dấu "=" xảy ra <=> x+3=0 và y-2=0 <=> x=-3 và y=2

Vậy ........

P < = 2018

Dấu "=" xảy ra <=> x+2=0 <=> x=-2

Vậy ...........

k mk nha

Đặt \(\sqrt{x^2+y^2}=c;\sqrt{y^2+z^2}=a;\sqrt{z^2+x^2}=b\)

Ta có:

\(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)

\(\ge\frac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\frac{y^2}{\sqrt{2\left(z^2+x^2\right)}}+\frac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)

\(=\frac{1}{2\sqrt{2}}\left(\frac{c^2+b^2-a^2}{a}+\frac{a^2+c^2-b^2}{b}+\frac{b^2+a^2-c^2}{c}\right)\)

\(\ge\frac{1}{2\sqrt{2}}\left(\frac{\left(2a+2b+2c\right)^2}{2\left(a+b+c\right)}-2018\right)=\frac{1009}{\sqrt{2}}\)

\(A=\left(x+1\right)^2-\left(2x-3\right)^2-15\)

\(A=\left(x+1\right)^2-\left(2x-3\right)^2-15\ge-15\)

\(A_{min}=-15\Leftrightarrow\orbr{\begin{cases}x+1=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{2}\end{cases}}}\)

P/s tham khảo nha

\(B=x^2-2xy+4y^2-2x-10y+2018\)

\(B=\left(x^2-2xy+y^2\right)-\left(2x-2y\right)+1+\left(3y^2-12y+12\right)+2005\)

\(B=\left(x-y\right)^2-2\left(x-y\right)+1+3\left(y-2\right)^2+2005\)

\(B=\left(x-y-1\right)^2+3\left(y-2\right)^2+2005\ge2005\)

VÌ \(\left(x-y-1\right)^2+3\left(y-2\right)^2\ge0\forall x;y\)

DẤU "="XẢY RA KHI Y=2;X=3