NGUYỄN MINH TÀI Ok bí thì cx đừng gắt,t giải đoạn đó cho
\(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
\(VT=\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\)
\(=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\)
\(\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)
\("="\Leftrightarrow\left(\sqrt{x-1}-2\right)\left(3-\sqrt{x-1}\right)\ge0\)
\(\Leftrightarrow2\le\sqrt{x-1}\le3\Leftrightarrow4\le x-1\le9\)
\(\Leftrightarrow5\le x\le10\)
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)}^2+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
Làm nốt nhé :v
