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Tìm x,y,z thỏa mãn x + y + z + 8 = 2 $\sqrt[]{x – 1}$ + 4 $\sqrt[]{ y – 2}$ + 6$\sqrt[]{z – 3}$
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Đáp án:
$x=2,y=6,z=12$
Giải thích các bước giải:
$x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}$
$x-1+2\sqrt{x-1}+1+y-2+4\sqrt{y-2}+4+z-3+6\sqrt{z-3}+9=0$
$(\sqrt{x-1}-1)^2+(\sqrt{y-2}-2)^2+(\sqrt{z-3}-3)^2=0$
vì:
$(\sqrt{x-1}-1)^2 ≥0$
$(\sqrt{y-2}-2)^2≥0$
$(\sqrt{z-3}-3)^2≥0$
⇒$(\sqrt{x-1}-1)^2+(\sqrt{y-2}-2)^2+(\sqrt{z-3}-3)^2≥0$
nên phương trình
⇔$\sqrt{x-1}=1$ ,$\sqrt{y-2}=2$,$\sqrt{z-3}=3$
⇔$x=2,y=6,z=12$