a) \(\left|3,5-x\right|=1,3\)
\(\Rightarrow\left[{}\begin{matrix}3,5-x=1,3\\3,5-x=-1,3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3,5-1,3\\x=3,5+1,3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2,2\\x=4,8\end{matrix}\right.\)
b) \(1,6-\left|x-0,2\right|=0,4\)
\(\Rightarrow\left|x-0,2\right|=1,2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,2=1,2\\x-0,2=-1,2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1,2+0,2\\x=-1,2+0,2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1,4\\x=-1\end{matrix}\right.\)
\(\left|3,5-x\right|=1,3\)
\(\Rightarrow\left[{}\begin{matrix}3,5-x=1,3\Rightarrow x=2,2\\3,5-x=-1,3\Rightarrow x=4,8\end{matrix}\right.\)
\(1,6-\left|x-0,2\right|=0,4\)
\(\Rightarrow\left|x-0,2\right|=1,2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,2=1,2\Rightarrow x=1,4\\x-0,2=-1,2\Rightarrow x=-1\end{matrix}\right.\)
\(\left|x-1,5\right|+\left|2,5-x\right|=0\)
\(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x-1,5\right|=0\Rightarrow x=1,5\\\left|2,5-x\right|=0\Rightarrow x=2,5\end{matrix}\right.\)
\(1,5\ne2,5\Rightarrow x\in\varnothing\)
c) \(\left|x-1,5\right|+\left|2,5-x\right|=0\)
Ta có : \(\left|x-1,5\right|\ge0\) với mọi \(x\)
\(\left|2,5-x\right|\ge0\) với mọi \(x\)
Nên \(\left|x-1,5\right|+\left|2,5-x\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)
Vậy không tìm được giá trị thõa mãn của \(x\)
Chúc học tốt !!!
