\(x^2+2y^2-2xy+y=0\) đề phải như thế này chứ

\(A=2x^2-2x+9-2xy+y^2\)

\(\Leftrightarrow A=\left(x^2-2x+1\right)+\left(x^2-2xy+y^2\right)+8\)

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

Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\forall x\\\left(x-y\right)^2\ge0\forall x;y\end{cases}}\)=> \(A=\left(x-1\right)^2+\left(x-y\right)^2+8\ge8\)

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

Vậy MinA = 8 <=> x = y = 1

\(2x^2+y^2+2x-2xy+5-4y=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}y-x-2=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

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

\(=3^2+2^2=13\)

bấn MT

cần gấp

a) =2x - 3 =0

     x = 3/2

b) (5x -1)² = 0

     5x - 1 =  0

       x = 1/5

c) = ( x +3)² = 0

        x+3  = 0

         x = -3

d) =(13+y)(13-y) = 0

        y = 13; -13

e) xem lại đề bài này

thank bạn nhiều lắm

a ) ( 2 x - 3 ) ^ 2 = 0

=> 2 x - 3           = 0

     2 x                = 3

        x                = 1,5

b ) 25 x ^ 2  - 10 x + 1 = 0

  (  5 x ) ^ 2   -  2 . 5 x + 1 ^ 2 = 0

  ( 5 x - 1 ) ^  2                       = 0

  5 x - 1                                  = 0

  5x                                        = 1

    x                                        = 0,2

c ) 6 x + 9 = - x ^ 2

   6 x + 9 + x ^ 2 = 0

  x ^ 2 + 2 . x . 3 + 3 ^ 2 = 0

  ( x + 3 ) ^ 2                  = 0

  x + 3                            = 0

        x                            = -3

d ) 169 - y ^ 2 = 0

            y  ^ 2  = 169

            y ^ 2   = 13 ^ 2

=> y     = 13