1/ \(\left(x-1\right)^4=16\left(x-1\right)^2\)
\(\Leftrightarrow\left(x-1\right)^4-16\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-16\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x-1\right)^2-16=0\Rightarrow\left(x-1\right)^2=16\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\\x=-3\end{matrix}\right.\)
Vậy==..
2/ \(\left|2x+1\right|+\left|x+8\right|=x\)
\(\Leftrightarrow\left|2x+1\right|=x-\left|x+8\right|\)
+) Với \(x\ge-\dfrac{1}{2}\) ta có:
\(2x+1=x-x-8\)
\(\Leftrightarrow2x+1=-8\Leftrightarrow2x=-9\Leftrightarrow x=-\dfrac{9}{2}\) (không t/m)
+) Với \(x< -\dfrac{1}{2}\) có:
\(2x+1=x+x+8\)
\(\Leftrightarrow2x+1=2x+8\Leftrightarrow0x=7\) (vô lí)
Vậy không có giá trị nào của x thỏa mãn đề bài
