a) \(\left|x-1\right|=2x-5\)
\(\Rightarrow\left|x-1\right|-2x=-5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1-2x=-5\left(đk:x-1\ge0\right)\\-\left(x-1\right)-2x=-5\left(đk:x-1< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x-1=-5\\-x+1-2x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-5+1\\-3x+1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-4\\-3x=-5-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-4\\-3x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(đk:x\ge1\right)\\x=2\left(đk:x< 1\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
b) \(||x+5|-4|=3\)
\(\Rightarrow\left[{}\begin{matrix}\left|x+5\right|-4=3\\\left|x+5\right|-4=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x+5\right|=3+4\\\left|x+5\right|=-3+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x+5\right|=7\\\left|x+5\right|=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+5=7\\x+5=-7\\x+5=1\\x+5=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=7-5\\x=-7-5\\x=1-5\\x=-1-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-12\\x=-4\\x=-6\end{matrix}\right.\)
Vậy \(x_1=-12;x_2=-6;x_3=-4;x_4=2\)
a, \(\left|x-1\right|=2x-5\)
\(\Rightarrow\left[{}\begin{matrix}x-1=2x-5\\x-1=-\left(2x-5\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy x = 4 hoặc x = 2
b, | \(\left|x+5\right|-4\) | = 3
\(\Rightarrow\left[{}\begin{matrix}\left|x+5\right|-4=3\\\left|x+5\right|-4=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left|x+5\right|=7\\\left|x+5\right|=1\end{matrix}\right.\)
+) \(\left|x+5\right|=7\Rightarrow\left[{}\begin{matrix}x+5=-7\\x+5=7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-12\\x=2\end{matrix}\right.\)
+) \(\left|x+5\right|=1\Rightarrow\left[{}\begin{matrix}x+5=1\\x+5=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-6\end{matrix}\right.\)
Vậy \(x\in\left\{-12;2;-4;-6\right\}\)
a).
nếu \(x\ge1\) thì : \(\left|x-1\right|=x-1\)
nếu \(x< 1\) thì: \(\left|x-1\right|=1-x\)
từ 2 ĐK trên, ta có:
\(\left[{}\begin{matrix}x-1=2x-5\left(với\: x\ge1\right)\\1-x=2x-5\left(với\:x< 1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-4\Rightarrow x=4\left(nhận\right)\\-3x=-6\Rightarrow x=2\left(loại\right)\end{matrix}\right.\)
vậy phương trình đã cho có tập nghiệm là S={4}
b).
\(\left|\left|x+5\right|-4\right|=3\\ \left(\left|x+5\right|-4\right)^2=9\)
\(nếu\:x\ge-5\:thì:\:\left|x+5\right|=x+5\\ nếu\:x< -5\:thì:\:\left|x+5\right|=-x-5\)
từ 2 ĐK trên, ta có:
\(\left[{}\begin{matrix}\left(x+5-4\right)^2=9\left(với\:x\ge-5\right)\\\left(-x-5-4\right)^2=9\left(với\:x< -5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\\\left[{}\begin{matrix}-x-9=3\\-x-9=-3\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-4\left(nhận\right)\\x=-12\left(nhận\right)\\x=-5\left(loại\right)\end{matrix}\right.\)
vậy phương trình đã cho có tập nghiệm là S={2;-4;-12}
a,\(\left|x-1\right|=2x-5\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=2x-5\\x-1=-\left(2x-5\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-2x=-5+1\\x-1=-2x+5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-x=-4\\x+2x=5+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\3x=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{2;4\right\}\)
b,\(\left|\left|x+5\right|-4\right|=3\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+5\right|-4=-3\\\left|x+5\right|-4=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+5\right|=1\\\left|x+5\right|=7\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+5=-1\\x+5=1\\x+5=-7\\x+5=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\x=-4\\x=-12\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{-12;-6;-4;2\right\}\)
Chúc bạn học tốt!!!
