\(\left(x-a+b\right)\left(x+2a-b-1\right)\le0\)(1)
\(\left|x+a-2\right|\le b+1\) (2)
Lời giải (khác)
\(b>-1\Leftrightarrow b+1>0\)
\(\left(2\right)\Leftrightarrow\left(x+a-2\right)^2\le\left(b+1\right)^2\)
\(\Leftrightarrow\left(x+a-2\right)^2-\left(b+1\right)^2\le0\)
\(\Leftrightarrow\left[\left(x+a-2\right)-\left(b+1\right)\right]\left[\left(x+a-2\right)+\left(b+1\right)\right]\le0\)
\(\Leftrightarrow\left(x+a-b-3\right)\left(x+a+b-1\right)\le0\)
Để \(\left(1\right)\Leftrightarrow\left(2\right)\Rightarrow\) a,b cần thỏa mãn :
\(\left[{}\begin{matrix}\left(I\right)\left\{{}\begin{matrix}-a+b=a-b-3\\2a-b-1=a+b-1\end{matrix}\right.\\\left(II\right)\left\{{}\begin{matrix}-a+b=a+b-1\\2a-b-1=a-b-3\end{matrix}\right.\end{matrix}\right.\)
(I)\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2a-2b=3\\a-2b=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=3\\b=\dfrac{3}{2}\end{matrix}\right.\)
(II)\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2a=-1\\a=-3\end{matrix}\right.\) vô No
Kết luận
Cặp a,b duy nhất thủa mãn là: (a,b)=(3,3/2)
