Question

[Nguyễn Bảo Minh]

Trả lời giúp mình nhé, Mình cần gấp, mai nộp bài rùi:

Answer 1

[Hoang Hung Quan]

Bàì 1:

Dễ thấy: \(\left\{{}\begin{matrix}\left(x+2\right)^2\ge0\forall x\\\left|y-3\right|\ge0\forall y\end{matrix}\right.\)

\(\Leftrightarrow\left(x+2\right)^2+\left|y-3\right|\ge0\)

Do đó để \(\left(x+2\right)^2+\left|y-3\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)^2=0\\\left|y-3\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\y-3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0-2\\y=0+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)

Vậy \(\left(x,y\right)=\left(-2;3\right)\)

Answer 2

[Aki Tsuki]

Bài 2:

\(\dfrac{-\left(x+1\right)}{4}=\dfrac{-9}{x+1}\Rightarrow-\left(x+1\right)\left(x+1\right)=-9\cdot4\)

\(\Rightarrow-\left[\left(x+1\right)^2\right]=-\left(6^2\right)\)

\(\Rightarrow x+1=6\Rightarrow x=5\)