Đáp án`+`Giải thích các bước giải:

a) Ta có:

`{((2x - 1/6)^2 ≥ 0),(\sqrt{3y + 12} ≥ 0):}`

`⇒ (2x - 1/6)^2 + \sqrt{3y + 12} ≥ 0`

Để (2x - 1/6)^2 + \sqrt{3y + 12} ≤ 0` thì

`{((2x - 1/6)^2 = 0),(\sqrt{3y + 12} = 0):}`

`⇒ `{(2x - 1/6 = 0),(3y + 12 = 0):}`

`⇒ (x;y) = (1/12; 4)`

b)Ta có:

`{((x + 2z)^2 ≥ 0),(\sqrt{x - 1} ≥ 0),(|2x - y| ≥ 0):}`

Để `(x + 2z)^2 + |2x - y| = -\sqrt{x - 1}`

`⇒ x = 2z; 2x = y; x = 1`

`⇒ (x;y;z) = (1;2;1/2)`

꧁$\color{#1c1c1c}{\text{G}}$$\color{#2a2043}{\text{o}}$$\color{#291063}
{\text{d}}$$\color{#53008}{\text{M}}$$\color{#6b28ac}
{\text{a}}$$\color{#8c4ebd}{\text{t}}$$\color{#B592D6}
{\text{h}}$$\color{#ceaedf}{\text{s}}$꧂