Ai giúp mik vs

Huhu ai giúp vs

Ta có:

\(A=x^2+y^2+xy-2x-4y+2016\\ =\left(x+\dfrac{y}{2}-1\right)^2+\dfrac{3}{2}\left(y-1\right)^2+\dfrac{4027}{2}\\ \ge\dfrac{4027}{2}\)

Dấu bằng xảy ra khi và chỉ khi: 

\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=1\end{matrix}\right.\)

A=x²-2x+1+y²-4y+4+2 = (x-1)²+(y-2)² + 2\(\ge\)2 Với mọi x, y

=> A_min = 2 đạt được khi x=1 và y=2

\(A=x^2-2x+y^2-4y-7=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)-12.\)

\(=\left(x-1\right)^2+\left(y-2\right)^2-12\)

Vì \(\left(x-1\right)^2+\left(y-2\right)^2\ge0\)nên \(\left(x-1\right)^2+\left(y-2\right)^2-12\ge-12\)

Vậy GTNN của A là -12 tại \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}}\)

a) A= 2x²-8x+10 = 2(x-2)²+2\(\ge\)2\(\Leftrightarrow\)x=2

Vậy MinA=2 \(\Leftrightarrow\)x=2

b) B= -(x-1)²-(2y+1)²+7 \(\le\)7

Dấu = xảy ra khi x=1 và y=\(\frac{-1}{2}\)

Vậy MaxB=7 ....

cảm ơn bạn nha

Min A=10<=>x=-1/2

Min B=2<=>x=1,y=2

Giups mik vs

\(E=2x^2+5y^2+x+4y+5\)

\(\Rightarrow E=2x^2+x+5y^2+4y+5\)

\(\Rightarrow E=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}-\dfrac{1}{16}\right)+5\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}-\dfrac{4}{25}\right)+5\)

\(\Rightarrow E=2\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)+5\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)+5-\dfrac{1}{8}-\dfrac{4}{5}\)

\(\Rightarrow E=2\left(x+\dfrac{1}{4}\right)^2+5\left(y+\dfrac{2}{5}\right)^2+\dfrac{163}{40}\)

mà \(\left\{{}\begin{matrix}2\left(x+\dfrac{1}{4}\right)^2\ge0,\forall x\\5\left(y+\dfrac{2}{5}\right)^2\ge0,\forall y\end{matrix}\right.\)

\(\Rightarrow E=2\left(x+\dfrac{1}{4}\right)^2+5\left(y+\dfrac{2}{5}\right)^2+\dfrac{163}{40}\ge\dfrac{163}{40}\)

\(\Rightarrow GTNN\left(E\right)=\dfrac{163}{40}\left(tạix=-\dfrac{1}{4};y=-\dfrac{2}{5}\right)\)

\(A=x^2+2x\left(y+1\right)+\left(y+1\right)^2-\left(y+1\right)^2+2y^2-4y+2028\)

\(=\left(x+y+1\right)^2-y^2-2x-1+2y^2-4y+2028\)

\(=\left(x+y+1\right)^2-6x+y^2+2027\)

\(=\left(x+y+1\right)+\left(y-3\right)^2+2018\ge2018\forall x;y\) (do...)

=> MinA = 2018 \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=3\end{matrix}\right.\)