Từ Phương trình ta suy ra:
\(\left|x-1\right|^{10}\le1\Leftrightarrow\left|x-1\right|\le1\Leftrightarrow-1\le x-1\le1\Leftrightarrow0\le x\le2\left(1\right)\)
\(\left|x-2\right|^{20}\le1\Leftrightarrow\left|x-2\right|\le1\Leftrightarrow-1\le x-2\le1\Leftrightarrow1\le x\le3\left(2\right)\)
Kết hợp \(\left(1\right)\left(2\right)\): \(1\le x\le2\)
\(\Rightarrow x-1\ge0\Leftrightarrow\left|x-1\right|=x-1\)
\(\Rightarrow x-2\le0\Leftrightarrow\left|x-2\right|=2-x\)
Mặt khác :
\(\left|x-1\right|\le1\Leftrightarrow\left|x-1\right|^{10}\le\left|x-1\right|\)
\(\left|x-2\right|\le1\Leftrightarrow\left|x-2\right|^{20}\le\left|x-2\right|\)
Ta có:
\(1=\left|x-1\right|^{10}+\left|x-2\right|^{10}\le\left|x-1\right|+\left|x-2\right|=\left(x-1\right)+\left(2-x\right)=1\)
Đã xảy ra dấu \(=\) nên suy ra 2 trường hợp:
\(\left|x-1\right|^{10}=\left|x-1\right|=x-1=1\Leftrightarrow x=2thỏa\left|x-2\right|^{20}=0\)
\(\left|x-2\right|^{20}=\left|x-2\right|=2-x=1\Leftrightarrow x=1thỏa\left|x-1\right|^{10}=0\)
