Question

Tìm x,y:

l3x+9l + (2y+4)²\(\le0\)

Answer 1

Vì 

\(\left|3x+9\right|\ge0\)

\(\left(2y+4\right)^2\ge0\)

\(\Rightarrow\left|3x+9\right|+\left(2y+4\right)\ge0\)

Để \(\left|3x+9\right|+\left(2y+4\right)^2\le0\Rightarrow\orbr{\begin{cases}\left|3x+9\right|=0\\\left(2y+4\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\y=-2\end{cases}}}\)

Vậy x = - 3; y = - 2

Answer 2

Ta có: 

 \(\left|3x+9\right|>\)hoặc bằng 0.\(\left(1\right)\)

  \(\left(2y+4\right)^2>=0\)\(\left(2\right)\)

Từ (1) và (2)\(\Rightarrow\left|3x+9\right|=0\); \(\left(2y+4\right)^2=0\)

  \(\left|3x+9\right|=0\)

\(\Rightarrow3x+9=0\)

\(\Rightarrow3x=-9\)

\(\Rightarrow x=-3\)

ta lại có:

\(\left(2y+4\right)^2=0\)

\(\Rightarrow2y+4=0\)

\(\Rightarrow2y=-4\)

\(\Rightarrow y=-2\)

Vậy \(x=-3;y=-2\).