Vì \(\hept{\begin{cases}\left(2x-1\right)^{2010}\ge0\\\left(y-\frac{2}{5}\right)^{2010}\ge0\\\left|x+y-z\right|\ge0\end{cases}\forall x,y,z}\)

\(\Rightarrow\left(2x-1\right)^{2010}+\left(y-\frac{2}{5}\right)^{2010}+\left|x+y-z\right|\ge0\)

Mà \(\left(2x-1\right)^{2010}+\left(y-\frac{2}{5}\right)^{2010}+\left|x+y-z\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x-1\right)^{2010}=0\\\left(y-\frac{2}{5}\right)^{2010}=0\\\left|x+y-z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}}\)

Vậy...

Ta có: \(\left(2x-1\right)^{2012}=\left(2x-1\right)^{2010}\)

=>\(\left(2x-1\right)^{2010}\left\lbrack\left(2x-1\right)^2-1\right\rbrack=0\)

=>\(\left(2x-1\right)\left(2x-1-1\right)\left(2x-1+1\right)=0\)

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

=>\(\left[\begin{array}{l}2x=0\\ 2x-1=0\\ 2x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac12\\ x=1\end{array}\right.\)

a) \(M=\left(2x-1\right)\left(2y-1\right)=4xy-2x-2y+1=4\left(xy\right)-2\left(x+y\right)+1\)

\(M=4.16-2.10+1=45\)

b) Ta có: 

\(\hept{\begin{cases}\left(x+2\right)^{2010}\ge0\\|y-\frac{1}{5}|\ge0\end{cases}}\left(\forall x,y\in R\right)\)

Khi đó \(N=\left(x+2\right)^{2010}+|y-\frac{1}{5}|-10\ge-10\)

Dấu "=" xảy ra khi x + 2 = 0 và y - 1/5 = 0 

Suy ra x = -2 và y = 1/5

( x - 1 )⁵ = 32

Mà 2⁵ = 32

=> x - 1 = 2

=> x = 2 + 1

=> x = 3

Vậy x = 3

( x - 1 )⁵ = 32                  y²⁰⁰ = y

( x - 1 )⁵ = 2⁵                    => y = 1

=> x - 1 = 2

           x = 2 + 1 

           x = 3

Vậy x = 3

hoc hang dang thuc dang nho chua

\(16\left(2x-1\right)^{2010}=\left(1-2x\right)^{2014}\)

\(\Rightarrow\)\(2^4\left(2x-1\right)^{2010}=\left(1-2x\right)^{2014}\) \(\Leftrightarrow\) \(2^4\left(2x-1\right)^{2010}=\left(1-2x\right)^{2010}.\left(1-2x\right)^4\)

\(\Rightarrow\)\(\hept{\begin{cases}2x-1=2\\1-2x=\left(-2\right)\end{cases}}\) ( Hai vế trái dấu )

\(\Rightarrow\) \(\hept{\begin{cases}x=\frac{3}{2}\\x=\frac{3}{2}\end{cases}}\)