Đáp án: $y=6\sqrt{2},x=8\sqrt{2}, z=9\sqrt{2}$
$y=-6\sqrt{2},x=-8\sqrt{2}, z=-9\sqrt{2}$
Giải thích các bước giải:
Ta có:
$|3x-4y|\ge 0,\quad\forall x,y$
$(3y-2z)^2\ge 0,\quad\forall y,z$
$\to |3x-4y|+(3y-2z)^2\ge 0,\quad\forall x,y,z$
Dấu = xảy ra khi $3x-4y=0, 3y-2z=0$
$\to 3x=4y, 2z=3y$
$\to x=\dfrac43y, z=\dfrac32y$
Mà $x^2+2y^2+z^2=434$
$\to (\dfrac43y)^2+2y^2+(\dfrac32y)^2=434$
$\to \frac{217}{36}y^2=434$
$\to y^2=72$
$\to y=\pm6\sqrt{2}$
Nếu $y=6\sqrt{2}\to x=8\sqrt{2}, z=9\sqrt{2}$
$y=-6\sqrt{2}\to x=-8\sqrt{2}, z=-9\sqrt{2}$
