Question

[ngo hoang khang]

Tìm x,y biết: x² + 4y² - 2x + 4y + 2 = 0

                     5x² + 3y² + z² - 4x + 6xy + 4z + 6 = 0

                     2x² + 2y² + z² +2xy + 2xz + 2yz + 2x + 4y + 5 = 0

Answer 1

[l҉o҉n҉g҉ d҉z҉]

Ta có : x² + 4y² - 2x + 4y + 2 = 0

<=> (x² - 2x + 1) + (4y² + 4y + 1) = 0

<=> (x - 1)² + (2x + 1)² = 0

Mà : \(\left(x-1\right)^2\ge0\forall x\)

        \(\left(2x+1\right)^2\ge0\forall x\)

Nên \(\orbr{\begin{cases}x-1=0\\2x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=-1\end{cases}}\)

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