\(\Leftrightarrow\left|x+1\right|+\left|1-x\right|< 4\)
\(\Leftrightarrow x+1< 4\) hay \(x+1< -4\) ; \(\Leftrightarrow1-x< 4\) hay \(1-x< -4\)
\(\Leftrightarrow x< 3\) hay \(x< -5\) ; \(\Leftrightarrow x< -3\) hay \(x< 5\)
Vậy : S = {x|x < 3; -5; -3; 5}
chưa học tới mà hăm bíc có đúng hăm:) (I'm lớp 8 :>)
\(\left|x+1\right|+\left|1-x\right|-4< 0\Leftrightarrow\left\{{}\begin{matrix}x+1+1-x-4< 0\left(khi\right)x+1\ge0\Leftrightarrow x\ge-1\left(1\right)\\-\left(x+1\right)+1-x-4< 0\left(khi\right)-x+1< 0\Leftrightarrow x< 1\left(2\right)\\x+1+1-x-4< 0\left(khi\right)1-x\ge0\Leftrightarrow x\ge1\left(3\right)\\x+1+-\left(1-x\right)-4< 0\left(khi\right)-1+x< 0\Leftrightarrow x< 1\left(4\right)\end{matrix}\right.\)
Giải pt (1) khi \(x\ge1\) :
\(x+1+1-x-4< 0\)
\(2-4< 0\)
=>\(x\ge1\)
giải pt (2) khi x < 1:
\(-\left(x+1\right)+1-x-4< 0\)
<=> \(-x-1+1-x-4< 0\)
<=> \(-2x-4< 0\)
<=> \(-2x< 4\)
<=> \(x>-\dfrac{4}{2}\)
=> \(x>-2\) ( nhận )
giải pt (3) khi \(x\ge1\) :
\(x+1+1-x-4< 0\)
<=> \(2-4< 0\)
=> \(x\ge1\)
giải pt (4) khi x < 1:
\(x+1+-\left(1-x\right)-4< 0\)
<=> \(x+1-\left(1-x\right)-4< 0\)
<=> \(x+1-1+x-4< 0\)
<=> \(2x-4< 0\)
<=> \(2x< 4\)
<=> \(x< \dfrac{4}{2}\)
=> \(x< 2\) ( loại )
Vậy \(S=\left\{x|x\ge1;-2\right\}\)