Tìm các số x;y;z thỏa mãn biết : `x+y+z+8=` 2 √(x-1) +4 √(y-2) +6 √(z-3)

Đáp án:

$\left\{\begin{array}{l} x=2\\ y=6 \\ z=12\end{array} \right..$

Giải thích các bước giải:

$x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}(x \ge 1; y \ge 2;  z \ge 3)\\ \Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\\ \Leftrightarrow (x-1)-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+(z-3)-6\sqrt{z-3}+9=0\\ \Leftrightarrow \left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\\ \text{Do } \left(\sqrt{x-1}-1\right)^2 \ge 0 \ \forall \ x \ge 1\\ \left(\sqrt{y-2}-2\right)^2 \ge 0 \ \forall \ y \ge 2\\ \left(\sqrt{z-3}-3\right)^2  \ge 0 \ \forall \ z \ge 3\\ \Rightarrow \left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2 \ge 0 \ \forall \ x \ge 1; y \ge 2;  z \ge 3$

Dấu "=" xảy ra

$\Leftrightarrow \left\{\begin{array}{l} \sqrt{x-1}-1=0\\ \sqrt{y-2}-2=0 \\\sqrt{z-3}-3=0\end{array} \right.\Leftrightarrow \left\{\begin{array}{l} \sqrt{x-1}=1\\ \sqrt{y-2}=2 \\\sqrt{z-3}=3\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x-1=1\\ y-2=4 \\ z-3=9\end{array} \right.\Leftrightarrow \left\{\begin{array}{l} x=2\\ y=6 \\ z=12\end{array} \right.$