giúp mình vs

\(4x^2+2y^2+2y-4xy+1=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(y^2+2y+1\right)=0\)

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

Vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0;\forall;x,y\\\left(y+1\right)^2\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(2x-y\right)^2+\left(y+1\right)^2\ge0;\forall x,y\)

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

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

\(\Leftrightarrow\hept{\begin{cases}2x+1=0\\y=-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=-1\end{cases}}\)

Vậy ...

sai đè nhá

giúp mới nà

Gọi 1/4 số a là 0,25 . Ta có :

                   a . 3 - a . 0,25 = 147,07

                   a . (3 - 0,25) = 147,07 ( 1 số nhân 1 hiệu )

                      a . 2,75 = 147,07

                         a = 147,07 : 2,75

                          a = 53,48

bạn nói rõ 1 chút được ko

\(\hept{\begin{cases}x^2+2y-4x=0\\4x^2-4xy^2+y^4-2y+4=0\end{cases}}\)

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

\(\Leftrightarrow2y+2y-4x=4x^2-x^2-4xy^2+y^4+4\)

\(\Leftrightarrow4y-4x=3x^2-y^2\left(4x-y^2\right)+4\)

\(\Leftrightarrow4y-4x-4=3x^2-y^2\left(4x-y^2\right)\)

\(\Leftrightarrow4\left(y-x-1\right)=3x^2-y^2\left(4x-y^2\right)\)

\(\Leftrightarrow4\left(y-x-1\right)=0\)

\(\Leftrightarrow y-x-1=\frac{0}{4}\)

\(\Leftrightarrow y-x-1=0\)

\(\Leftrightarrow y-x=0+1\)

\(\Leftrightarrow y-x=1\)

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

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

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+2\left(2x-y\right)+1+4=0\)

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

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

Với mọi x, y ta có :

\(\left(2x-y+1\right)^2\ge0\Leftrightarrow\left(2x-y+1\right)^2+4>0\)

\(\Leftrightarrow pt\) vô nghiệm