Ta có \(\hept{\begin{cases}\left(3x-5\right)^{2008}\ge0\\\left(y^2-1\right)^{2010}\ge0\\\left(x-z\right)^{2012}\ge0\end{cases}}\)mà \(\left(3x-5\right)^{2008}+\left(y^2-1\right)^{2010}+\left(x-z\right)^{2012}=0\)

\(\Rightarrow\hept{\begin{cases}\left(3x-5\right)^{2008}=0\\\left(y^2-1\right)^{2010}=0\\\left(x-z\right)^{2012}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}3x-5=0\\y^2-1=0\\x-z=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=1;-1\\z=x=\frac{5}{3}\end{cases}}\)

\(a,\Leftrightarrow y^{200}-y=y\left(y^{199}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y^{199}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=1\end{matrix}\right.\)

Vậy ..

\(b,\Leftrightarrow y^{2010}-y^{2008}=y^{2008}\left(y^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y^{2008}=0\\y^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y=1\\y=-1\end{matrix}\right.\)

Vậy ...

\(c,\Leftrightarrow\left(2y-1\right)^{50}-\left(2y-1\right)=\left(2y-1\right)\left(\left(2y-1\right)^{49}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2y-1=0\\\left(2y-1\right)^{49}=1\end{matrix}\right.\)

\(\Leftrightarrow y=\dfrac{1}{2}\)

Vậy ..

\(d,\Leftrightarrow\left(\dfrac{y}{3}-5\right)^{2008}\left(\left(\dfrac{y}{3}-5\right)^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(\dfrac{y}{3}-5\right)^{2008}=0\\\left(\dfrac{y}{3}-5\right)^2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{y}{3}-5=0\\\dfrac{y}{3}-5=1\\\dfrac{y}{3}-5=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y=15\\y=18\\y=12\end{matrix}\right.\)

Vậy ..

Nhanh lên nhé mình xin các bạn đấy

(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

VÌ \(\left(x-1\right)^{2012}\ge0\)

\(\left(y-2\right)^{2010}\ge0\)

\(\left(x-z\right)^{2008}\ge0\)

nên dấu \(=\)xảy ra khi \(\hept{\begin{cases}x=z\\x=1\\y=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=z=1\\y=2\end{cases}}}\)

CM : ( 3x - 2y )^2010 = 0 ,  / 5y - 6z /^2011 = 0 

=> 3x - 2y = 0    ,  5y - 6z = 0 

=> 3x = 2y     , 5y = 6z 

=> x/2 = y/3    , y/6 = z/5 

=> x/4 = y/6    , y/6 =z/5 

=> x/4 = y/6 = z/5 

=> 2x/ 8 , 5y/30 , 3z/15

Áp dụng tính chất DTSBN , ta có : 

2x/8 = 5y /30 = 3z / 15 = 2x - 5y + 3z / 8 - 30 + 15 = 54/-7 = -54 /7 

Rồi tính ra là xong