Ai giúp mik vs

Huhu ai giúp vs

\(A=\left(2x-y+1\right)^2+\left(x-3\right)^2-4y+2007\)

\(=4x^2+y^2+1-4xy+4x-2y+x^2-6x+9-4y+2007\)

\(=5x^2-4xy-2x-6y+y^2+2017\)

\(=\left[y^2-2y\left(2x+3\right)+\left(2x+3\right)^2\right]+\left(x^2-14x+49\right)+1959\)

\(=\left(y-2x-3\right)^2+\left(x-7\right)^2+1959\ge1959\)

\(minA=1959\Leftrightarrow\) \(\left\{{}\begin{matrix}x=7\\y=17\end{matrix}\right.\)

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=> giúp mình được ko

Giups mik vs

a) -( x-y)² - (x-1)² -2 

GTLN = -2

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) 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

x² - 2x + y² - 4y + 7 = (x² - 2x + 1) + ( y² - 4y + 4) + 2 = (x - 1)² + (y - 2)² + 2

Vì (x - 1)² ≥ 0 \(\forall\)x

    (y - 2)² ≥ 0 \(\forall\)x

=> (x - 1)² + (y - 2)² ≥ 0 \(\forall\)x

=> (x - 1)² + (y - 2)² + 2  ≥ 2 

Dấu " = " xảy ra <=> \(\hept{\begin{cases}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-1=0\\y-2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\end{cases}}\)

Vậy GTNN của x² - 2x + y² - 4y +7 = 2 khi x = 1; y = 2

Đặt \(A=x^2-2x+y^2-4y+7\)

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

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

Vì \(\left(x-1\right)^2\ge0\forall x\); \(\left(y-2\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)

hay \(A\ge2\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)

Vậy \(minA=2\)\(\Leftrightarrow\hept{\begin{cases}x=1\\y=2\end{cases}}\)

a) \(A=1-8x-x^2=-\left(x^2+8x+16\right)+17=-\left(x-4\right)^2+17\le17\)

\(ĐTXR\Leftrightarrow x=4\)

b) \(B=5-2x+x^2=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)

\(ĐTXR\Leftrightarrow x=1\)

c) \(C=x^2+4y^2-6x+8y-2021=\left(x^2-6y+9\right)+\left(4y^2+8y+4\right)-2034=\left(x-3\right)^2+\left(2y+2\right)^2-2034\ge-2034\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

a: Ta có: \(A=-x^2-8x+1\)

\(=-\left(x^2+8x-1\right)\)

\(=-\left(x^2+8x+16-17\right)\)

\(=-\left(x+4\right)^2+17\le17\forall x\)

Dấu '=' xảy ra khi x=-4

b: Ta có: \(x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi x=1