Ta có: \(\left|x+1\right|+\left|y-2\right|=0\)
\(\Rightarrow\left[\begin{matrix}\left|x+1\right|=0\\\left|y-2\right|=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Thay \(x=-1;y=2\) vào biểu thức C ta có:
\(C=-2+\frac{-1-6}{1+2}=-2+\frac{-7}{3}=\frac{-13}{3}\)
Vậy \(C=\frac{-13}{3}\)
Xét :\(\left\{\begin{matrix}\left|x+1\right|\ge0\\\left|y-2\right|\ge0\end{matrix}\right.\)
Để \(\left|x+1\right|+\left|y-2\right|=0\)
Thì: \(\left\{\begin{matrix}\left|x+1\right|=0\\\left|y-2\right|=0\end{matrix}\right.\)
\(\left\{\begin{matrix}\left|x+1\right|=0\\\left|y-2\right|=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Thay vào C:
\(C=2.\left(-1\right)+\dfrac{\left(-1\right)-3.2}{\left(-1\right)^2+2}=\dfrac{-13}{3}\)
Vậy \(C=\dfrac{-13}{3}\)
Giải:
Do \(\left|x+1\right|\ge0\) với mọi \(x\) và \(\left|y-z\right|\ge0\) với mọi \(y\)
Mà \(\left|x+1\right|+\left|y-z\right|=0\)
\(\Rightarrow\left|x+1\right|=0\) và \(\left|y-z\right|=0\)
\(\Rightarrow x=-1\) và \(y=2\)
\(C=2.\left(-1\right)+\dfrac{\left(-1\right)-3.2}{\left(-1\right)^2+3}\)
\(C=2+\dfrac{-7}{4}\)
\(C=\dfrac{-6}{4}+\dfrac{-7}{4}\)
\(C=\dfrac{-13}{4}\)