Question

\(\text{(2x +1)^2+| y-1,2| =0}\)

 

Answer 1

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

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

Answer 2

Vì (2x+1)²\(\ge0\) và \(\left|y-1,2\right|\ge0\)

Mà: \(\left(2x+1\right)^2+\left|y-1,2\right|=0\)

Nên: \(\begin{cases}2x+1=0\\y-1,2=0\end{cases}\Rightarrow\begin{cases}2x=-1\\y=1,2\end{cases}\Rightarrow}\begin{cases}x=-0,5\\y=1,2\end{cases}}\)Vậy x=-0,5; y=1,2

Answer 3

Vì \(\left(2x+1\right)^2\ge0,\left|y-1,2\right|\ge0\)

\(\Rightarrow\left(2x+1\right)^2+\left|y-1,2\right|\ge0\)

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

\(\Leftrightarrow2x+1=0,y-1,2=0\)

\(\Leftrightarrow2x=-1,y=1,2\)

\(\Leftrightarrow x=\frac{-1}{2},y=1,2\)