Question

Cho x,y,z thỏa mãn đẳng thức

x+y+z+8=2\(\sqrt{ }\) x-1 + 4\(\sqrt{ }\) y-1 + 6\(\sqrt{ }\) z-3

Answer 1

\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\) ( Sửa đề )

\(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\left(x\ge1;y\ge2;z\ge3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\left(TM\right)\\y=6\left(TM\right)\\z=12\left(TM\right)\end{matrix}\right.\)

KL..........

Answer 2

bn có ghi sai đề k v