+) ta có : \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=3\Leftrightarrow\left|x-1\right|+\left|x-2\right|=3\)
\(\)th1: \(x< 1\) \(\Rightarrow pt\Leftrightarrow3-2x=3\Leftrightarrow x=0\left(tmđk\right)\)
th2: \(1\le x< 2\) \(\Rightarrow pt\Leftrightarrow x-1+2-x=3\Leftrightarrow1=3\left(vôlí\right)\)
th3: \(x\ge2\) \(\Rightarrow pt\Leftrightarrow2x-3=3\Leftrightarrow x=3\left(tmđk\right)\)
vậy \(x=0;x=3\)
+) ở đây : https://hoc24.vn/hoi-dap/question/633500.html
\(\Leftrightarrow x+y+z+8-2\sqrt{x-1}+4\sqrt{y-2}+\sqrt{z-3}\ge0\)
\(\Leftrightarrow\left(x-1-2\sqrt{x-1}+1\right)+\left(y-2-4\sqrt{y-2}+4\right)+\left(z-3-6\sqrt{z-3}+9\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2\ge0\)
So ez, Tuấn ML