\(a,4x^2+9y^2+4x-24y+17=0\)

\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)

\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)

\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)

jup tui

x=2

y=-2

z=-2

a, Tìm GTNN

\(A=2x^2+y^2+2xy-8x+2028\)

\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+16\right)+2012\)

\(=\left(x+y\right)^2+\left(x-4\right)^2+2012\)

Ta có :

\(\left(x+y\right)^2\ge0\) với mọi x

\(\left(x-4\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x+y\right)^2+\left(x-4\right)^2+2012\ge2012\)

Dấu = xảy ra

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-4\right)^2=0\\\left(x+y\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-4=0\\x+y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-4\end{matrix}\right.\)

Vậy \(Min_A=2012\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-4\end{matrix}\right.\)

A=2x²+y²+2xy-8x+2028=(x²+2xy+y²)+(x²-8x+16)+2012=(x+y)²+(x-4)²+2012

Vì (x+y)^{2\(\ge\)0\(\forall\)x,y}

(x-4)^{2\(\ge0\forall x\)}

^=>(x+y)²+(x-4)²\(\ge0\)

=>(x+y)²+(x-4)²+2012\(\ge2012\forall x,y\)

Đạt được khi và chỉ khi:

\(\left\{{}\begin{matrix}x-4=0\rightarrow x=4\\x+y=0\rightarrow y=-4\end{matrix}\right.\)

Vậy Amin=2012<=>x=4,y=-4

a) A=2x²+y²+2xy-8x+2028

=(x²+2xy+y²)+(x²-8x+16)+2012

=(x+y)²+(x-4)²+2012

do (x+y) ²≥ 0 ∀x;y

(x-4)²≥ 0 ∀x

=> (x+y)²+(x-4)² ≥ 0

=> (x+y)²+(x-4)²+2012 ≥ 2012

=> A≥2012

vậy GTNN A=2012 khi \(\left[{}\begin{matrix}x+y=0\\x-4=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}y=-4\\x=4\end{matrix}\right.\)

a)\(x^2+2y^2+2xy-2y+1=0\)

\(\Leftrightarrow x^2+2xy+y^2+y^2-2y+1=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}y-1=0\\x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=-y=-1\end{cases}}\)

Vậy x=-1 y=1

a)  \(x^2+2y^2+2xy-2y+1=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+y=0\\y-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\y=1\end{cases}\Rightarrow}x=-1;y=1}\)

b) \(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)

\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(3x^2+6xy+3y^2\right)+\left(z^2+4z+4\right)=0\)

\(\Leftrightarrow2.\left(x-1\right)^2+3.\left(x+y\right)^2+\left(z+2\right)^2=0\)

\(\Rightarrow\)  \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)

           \(\left(x+y\right)^2=0\Rightarrow x+y=0\Rightarrow y=-x=-1\) 

            \(\left(z+2\right)^2=0\Rightarrow z+2=0\Rightarrow z=-2\)

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giúp mk đi. Mk đag cần gấp