x=5/2,y=-4/3

Vì \(\left(2x-5\right)^{2016}\ge0\forall x;\left(3y+4\right)^{2020}\ge0\forall y\)

\(\Rightarrow\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\ge0\)

Mà đề lại cho \(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\le0\)

Nên \(\hept{\begin{cases}\left(2x-5\right)^{2016}=0\\\left(3y+4\right)^{2020}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}}\)

Vậy ..........

vì (2x-5)²⁰¹⁶  và (3y+4)²⁰²⁰ >hoặc=0 với mọi x

=>2x-5=3y+4=0

=>x=2/5;y=-4/3

Vì: \(\left(2x-5\right)^{2016}\ge0;\left(3y+4\right)^{2020}\ge0\)

Nên:  \(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}\le0\)

\(\Leftrightarrow\left(2x-5\right)^{2016}+\left(3y+4\right)^{2020}=0\)

\(\Leftrightarrow\begin{cases}2x-5=0\\3y+4=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}\)

Ta có: \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\left(\forall x\right)\\\left(3y+4\right)^{2020}\ge0\left(\forall y\right)\end{cases}}\Rightarrow\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\left(\forall x,y\right)\)

Mà \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\left(\forall x,y\right)\)

\(\Rightarrow\hept{\begin{cases}\left(2x-5\right)^{2018}=0\\\left(3y+4\right)^{2020}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

Khi đó thay vào ta được: 

\(M+5\cdot\left(\frac{5}{2}\right)^2-2\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)=6\cdot\left(\frac{5}{2}\right)^2+9\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(\Leftrightarrow M+\frac{455}{12}=\frac{103}{18}\)

\(\Rightarrow M=-\frac{1159}{36}\)

1. Vì \(\left(x+6\right)^2\ge0\forall x\); \(\left|y-\frac{1}{2}\right|\ge0\forall y\); \(\left|x+y+z\right|\ge0\forall x,y,z\)

\(\Rightarrow\left(x+6\right)^2+\left|y-\frac{1}{2}\right|+\left|x+y+z\right|\ge0\)

mà \(\left(x+6\right)^2+\left|y-\frac{1}{2}\right|+\left|x+y+z\right|\le0\)( đề bài )

\(\Rightarrow\left(x+6\right)^2+\left|y-\frac{1}{2}\right|+\left|x+y+z\right|=0\)\(\Leftrightarrow\hept{\begin{cases}x+6=0\\y-\frac{1}{2}=0\\x+y+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-6\\y=\frac{1}{2}\\-6+\frac{1}{2}+z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-6\\y=\frac{1}{2}\\z=\frac{11}{2}\end{cases}}\)

Vậy \(x=-6\); \(y=\frac{1}{2}\); \(z=\frac{11}{2}\)

2. \(B=\left|x-2016\right|+\left|x-2018\right|=\left|x-2016\right|+\left|2018-x\right|\ge\left|x-2016+2018-x\right|=\left|2\right|=2\)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-2016\right)\left(2018-x\right)\ge0\)

TH1: \(\hept{\begin{cases}x-2016< 0\\2018-x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 2016\\2018< x\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 2016\\x>2018\end{cases}}\)( vô lý )

TH2: \(\hept{\begin{cases}x-2016\ge0\\2018-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2016\\2018\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2016\\x\le2018\end{cases}}\Leftrightarrow2016\le x\le2018\)( thoả mãn )

Vậy \(minB=2\Leftrightarrow2016\le x\le2018\)

Theo đề bài, ta có:

x(x + y + z) = -5; y(x + y + z) = 9; z(x + y + z) = 5

=> (x + y + z)(x + y + z) = -5 + 9 + 5 = 9

=> (x + y + z)² = 9

=> x + y + z \(\in\){3; -3}

Với x + y + z = 3, ta có:

   x = -5 : 3 = \(\frac{-5}{3}\)

   y = 9 : 3 = 3

   z = 5 : 3 = \(\frac{5}{3}\)

Với x + y + z = -3, ta có:

   x = -5 : (-3) = \(\frac{5}{3}\)

   y = 9 : (-3) = -3

   z = 5 : (-3) = \(\frac{-5}{3}\)

Vậy x = \(\frac{-5}{3}\); y = 3 ; z = \(\frac{5}{3}\) hoặc x = \(\frac{5}{3}\); y = -3 ; z = \(\frac{-5}{3}\).

x = 2,5

y = \(\frac{4}{3}\)

\(\left(2x-5\right)^{2000}\ge0;\left(3y-4\right)^{2002}\ge0\)

Mà \(\left(2x-5\right)^{2000}+\left(3y-4\right)^{2002}\le0\)

suy ra \(\left(2x-5\right)^{2000}+\left(3y-4\right)^{2002}=0\)

\(\Leftrightarrow\) (2x - 5)²⁰⁰⁰ = 0 và (3y - 4)²⁰⁰² = 0

\(\Leftrightarrow\) 2x - 5 = 0 và 3y - 4 = 0

\(\Leftrightarrow\) 2x = 5 và 3y = 4

\(\Leftrightarrow\) x = \(\frac{5}{2}\) và y = \(\frac{4}{3}\)

(2x-5)^2008 > 0

(3y+4)^2010 > 0

=>(2x-5)^2008+(3y+4)^2010>0

mà theo đề:(2x-5)^2008+(3y+4)^2010 < 0

=>(2x-5)^2008=(3y+4)^2010=0

+)(2x-5)^2008=0=>2x=5=>x=5/2

+)(3y+4)^2010=0=>3y=-4=>y=-4/3

Vậy...

vì 2008và 2010 chẵn nên (2x-5)^2008 và(3y+4)^2010> hoac = 0Vậy=0

x=5/2 và y =-4/3