Đáp án: $T=2012$
Giải thích các bước giải:
$ĐK: x≥34;y≥21;z≥4$
Ta có:
$x+y+z=2\sqrt{x-34}+4\sqrt{y-21}+6\sqrt{z-4}+45$
$⇔x+y+z-2\sqrt{x-34}-4\sqrt{y-21}-6\sqrt{z-4}-45=0$
$⇔[(x-34)-2\sqrt{x-34}+1]+[(y-21)-4\sqrt{y-21}+4]+[(z-4)-6\sqrt{z-4}+9]=0$
$⇔(\sqrt{x-34}-1)^2+(\sqrt{y-21}-2)^2+(\sqrt{z-4}-3)^2=0$
Do $(\sqrt{x-34}-1)^2≥0∀x$
$(\sqrt{y-21}-2)^2≥0∀y$
$(\sqrt{z-4}-3)^2≥0∀z$
$⇒(\sqrt{x-34}-1)^2+(\sqrt{y-21}-2)^2+(\sqrt{z-4}-3)^2≥0∀x;y;z$
Dấu bằng xảy ra
$⇔(\sqrt{x-34}-1)^2=(\sqrt{y-21}-2)^2=(\sqrt{z-4}-3)^2=0$
Ta có:
$(\sqrt{x-34}-1)^2=0⇔\sqrt{x-34}-1=0$
$⇔\sqrt{x-34}=1⇔x-34=1⇔x=35$ (thỏa mãn)
$(\sqrt{y-21}-2)^2=0⇔\sqrt{y-21}-2=0$
$⇔\sqrt{y-21}=2⇔y-21=4⇔y=25$ (thỏa mãn)
$(\sqrt{z-4}-3)^2=0⇔\sqrt{z-4}-3=0$
$⇔\sqrt{z-4}=3⇔z-4=9⇔z=13$ (thỏa mãn)
Ta có: $T=x^2+y^2+z^2-7=35^2+25^2+13^2-7=2012$
