Đề Bài : Tìm x; y; z \(\in\) Q , biết :
a) \(\left|x+\frac{3}{4}\right|\) + \(\left|y-\frac{1}{5}\right|\)+\(\left|x+y+z\right|\) = 0
b) \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{3}\right|+\left|z+\frac{1}{2}\right|=0\)
c) |x| +x = 0
d) \(\left|x+1\right|+\left|x^2-1\right|=0\)
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Đề Bài : Tìm x; y; z \(\in\) Q , biết :
a) \(\left|x+\frac{3}{4}\right|\) + \(\left|y-\frac{1}{5}\right|\)+\(\left|x+y+z\right|\) = 0
b) \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{3}\right|+\left|z+\frac{1}{2}\right|=0\)
c) |x| +x = 0
d) \(\left|x+1\right|+\left|x^2-1\right|=0\)
a) Ta có: \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|\ge0\)
Mà \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
\(\Rightarrow\left[\begin{matrix}\left|x+\frac{3}{4}\right|=0\\\left|x-\frac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x+\frac{3}{4}=0\\y-\frac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=\frac{-3}{4}\\y=\frac{1}{5}\\z=0-\frac{-3}{4}-\frac{1}{5}=\frac{11}{20}\end{matrix}\right.\)
Vậy \(x=\frac{-3}{4};y=\frac{1}{5};z=\frac{11}{20}\)
b) \(\left|x+\frac{3}{4}\right|+\left|y-\frac{2}{3}\right|+\left|z-\frac{1}{2}\right|=0\)
\(\Rightarrow\left[\begin{matrix}\left|x+\frac{3}{4}\right|=0\\\left|y-\frac{2}{3}\right|=0\\z+\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x+\frac{3}{4}=0\\y-\frac{2}{3}=0\\z+\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=\frac{-3}{4}\\y=\frac{2}{3}\\z=\frac{-1}{2}\end{matrix}\right.\)
Vậy \(x=\frac{-3}{4};y=\frac{2}{3};z=\frac{-1}{2}\)
d) \(\left|x+1\right|+\left|x^2-1\right|=0\)
\(\Rightarrow\left[\begin{matrix}\left|x+1\right|=0\\\left|x^2-1\right|=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x+1=0\\x^2-1=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=-1\\x=\pm1\end{matrix}\right.\)
Vậy \(x\in\left\{1;-1\right\}\)
thiếu phần c) rồi bạn ơi