Ta có: (x + 2y) + (2x - y)i = 6 + 7i

Vậy: T = 4 + 1 = 5

Chọn B 

\(x^3+x\ge2\sqrt{x^4}=2x^2\)

Tương tự:

\(y^3+y\ge2y^2\)

\(z^3+z\ge2z^2\)

Cộng vế:

\(x^3+y^3+z^3+x+y+z\ge2\left(x^2+y^2+z^2\right)=6\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Chọn D 

Ta có: x(3 + 5i) - y(1 + 2i) = 9 + 16i <=> (3x - y) + (5x - 2y) = 9 + 16i

Vậy: T = |x - y| = 5

tks mn

Ta có: \(\left(x-1\right)^2\ge0\forall x\)

\(\left|3y-1\right|\ge0\forall y\)

\(\left|z+2\right|\ge0\forall z\)

Do đó: \(\left(x-1\right)^2+\left|3y-1\right|+\left|z+2\right|\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left(x,y,z\right)=\left(1;\dfrac{1}{3};-2\right)\)

\(xy+x+1=3y\Rightarrow x+\dfrac{1}{y}+\dfrac{x}{y}=3\)

Ta có:

\(x^3+1+1\ge3x\)

\(\dfrac{1}{y^3}+1+1\ge\dfrac{3}{y}\)

\(x^3+\dfrac{1}{y^3}+1\ge\dfrac{3x}{y}\)

Cộng vế:

\(2\left(x^3+\dfrac{1}{y^3}\right)+5\ge3\left(x+\dfrac{1}{y}+\dfrac{x}{y}\right)=9\)

\(\Rightarrow x^3+\dfrac{1}{y^3}\ge2\)

\(\Rightarrow x^3y^3+1\ge2y^3\) (đpcm)

Dấu "=" xảy ra khi \(x=y=1\)

`(x-1)^2>=0`

`|3y-1|>=0`

`|z+2|>=0`

`=>(x-1)^2+|3y-1|+|z+2|>=0`

Mà đề bài cho =0

`=>{(x-1=0),(3y-1=0),(z+2=0):}`

`=>{(x=1),(y=1/3),(z=-2):}`

Vậy `x=1` và `y=1/3` và `z=-2`

Ta có: \(\left(x-1\right)^2\ge0\forall x\)

\(\left|3y-1\right|\ge0\forall y\)

\(\left|z+2\right|\ge0\forall z\)

Do đó: \(\left(x-1\right)^2+\left|3y-1\right|+\left|z+2\right|\ge0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\3y-1=0\\z+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{3}\\z=-2\end{matrix}\right.\)