Ta có : \(x^2+y^2\ge2xy\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)

\(\Leftrightarrow x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

Áp dụng vào bài toán có :

\(P\le\frac{x+y}{\frac{\left(x+y\right)^2}{2}}+\frac{y+z}{\frac{\left(y+z\right)^2}{2}}+\frac{z+x}{\frac{\left(z+x\right)^2}{2}}\) \(=\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}=\frac{1}{2}\left(\frac{4}{x+y}+\frac{4}{y+z}+\frac{4}{z+x}\right)\)

Áp dụng BĐT Svacxo ta có :

\(\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\), \(\frac{4}{y+z}\le\frac{1}{y}+\frac{1}{z}\), \(\frac{4}{z+x}\le\frac{1}{z}+\frac{1}{x}\)

Do đó : \(P\le\frac{1}{2}\left[2.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right]=2016\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{672}\)

P/s : Dấu "=" không chắc lắm :))

thanks bạn mình hiểu sương sương rồi:))

Chọn đáp án D

Ta có: z 1 = 1 - 2 i ⇒ z 1 = 1 + 2 i  

Khi đó

Vậy  x ; y = 6 ; 4

\(z^2=xy-16=\left(8-y\right)y-16=-y^2+8y-16\\ \Leftrightarrow z^2+\left(y-4\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}z=0\\y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=0\\x=8-4=4\end{matrix}\right.\\ \Leftrightarrow P=x+z=0+4=4\)

Từ x + y  = 2 => x = 2 - y thay vào xy - z² = 1

Ta có: \(\left(2-y\right)y-z^2=1\)

<=> \(z^2+y^2-2y+1=0\)

<=> \(z ^2+\left(y-1\right)^2=0\)

<=> \(\left\{{}\begin{matrix}z=0\\y=1\end{matrix}\right.\) => x = 2 - 1 = 1

Vậy x = y = 1 và z = 0

a, \(8^3yz+12^2yz+6xyz+yz\)

\(=512yz+144yz+6xyz+yz\)

\(=yz\left(512+14+6x+1\right)\)

\(=yz\left(527+6x\right)\)

$---$

b, \(81x^4\left(z^2-y^2\right)-z^2+y^2\)

\(=81x^4\left(z^2-y^2\right)-\left(z^2-y^2\right)\)

\(=\left(z^2-y^2\right)\left(81x^4-1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left[\left(9x^2\right)^2-1^2\right]\)

\(=\left(z-y\right)\left(z+y\right)\left(9x^2-1\right)\left(9x^2+1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left[\left(3x\right)^2-1^2\right]\left(9x^2+1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left(3x-1\right)\left(3x+1\right)\left(9x^2+1\right)\)

$---$

c, \(\dfrac{x^3}{8}-\dfrac{y^3}{27}+\dfrac{x}{2}-\dfrac{y}{3}\)

\(=\left[\left(\dfrac{x}{2}\right)^3-\left(\dfrac{y}{3}\right)^3\right]+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)

\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}\right)+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)

\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}+1\right)\)

$---$

d, \(x^6+x^4+x^2y^2+y^4-y^6\)

\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)

\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)

\(=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)+\left(x^4+x^2y^2+y^4\right)\)

\(=\left(x^4+x^2y^2+y^4\right)\left(x^2-y^2+1\right)\)

$Toru$

Có: x²+y²+z²≥1/3 (x+y+z)²  =4/3

=> x²+y²+z² -3 >= 4/3 - 3 = -5/3

Dấu "=" xảy ra khi x=y=z=2/3