Đáp án: $(x,y,z)\in\{(\dfrac32\cdot \sqrt{\dfrac{102400}{521}}, \sqrt{\dfrac{102400}{521}}, \dfrac75\cdot \sqrt{\dfrac{102400}{521}}),(-\dfrac32\cdot \sqrt{\dfrac{102400}{521}}, -\sqrt{\dfrac{102400}{521}}, -\dfrac75\cdot \sqrt{\dfrac{102400}{521}})\}$
Giải thích các bước giải:
Ta có: $(2x-3y)^2\ge 0,\quad\forall x,y$
$|7y-5z|\ge 0,\quad\forall y,z$
$\to (2x-3y)^2+|7y-5z|\ge 0,\quad\forall x, y,z$
Dấu = xảy ra khi $2x-3y=0, 7y-5z=0$
$\to 2x=3y$ và $7y=5z$
$\to x=\dfrac32y, z=\dfrac75y$
Mà $x^2+y^2+z^2=1024$
$\to (\dfrac32y)^2+y^2+(\dfrac75y)^2=1024$
$\to \dfrac{521}{100}y^2=1024$
$\to y^2=\dfrac{102400}{521}$
$\to y=\pm\sqrt{\dfrac{102400}{521}}$
$\to (x,y,z)\in\{(\dfrac32\cdot \sqrt{\dfrac{102400}{521}}, \sqrt{\dfrac{102400}{521}}, \dfrac75\cdot \sqrt{\dfrac{102400}{521}}),(-\dfrac32\cdot \sqrt{\dfrac{102400}{521}}, -\sqrt{\dfrac{102400}{521}}, -\dfrac75\cdot \sqrt{\dfrac{102400}{521}})\}$
