a) \(\left(2y-1\right)^{1000}-\left(3+y\right)^{1000}=0\)

\(\Rightarrow\left(2y-1\right)^{1000}=\left(3+y\right)^{1000}\)

\(\Rightarrow2y-1=3+y\)

\(2y-y=3+1\)

\(y=4\)

b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\left(x-\frac{2}{9}\right)^3=\left(\left(\frac{2}{3}\right)^2\right)^3\)

\(\Rightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2\)

\(x-\frac{2}{9}=\frac{4}{9}\)

\(x=\frac{2}{3}\)

c) \(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\left(\left(2x-1\right)^3\right)^2=\left(\left(2x-1\right)^4\right)^2\)

\(\Rightarrow\left(2x-1\right)^3=\left(2x-1\right)^4\)

\(8x^3-1=16x^4-1\)

\(16x^4-8x^3=0\)

\(8x^3\left(2x-1\right)=0\)

Nếu \(8x^3=0\) thì \(x^3=0\Rightarrow x=0\)

Nếu \(2x-1=0\)thì \(2x=1\Rightarrow x=\frac{1}{2}\)

Vậy x=0 và x=1/2

y=816

ta có:\(\left|x-1000\right|+\left|x-2019\right|=\left|-x+1000\right|+\left|x-2019\right|\)

\(\ge\left|-x+1000+x-2019\right|=1019\)

dấu = xảy ra khi \(\left(-x+1000\right).\left(x-2019\right)\ge0\)

\(\Rightarrow1000\le x\le2019\)

\(\hept{\begin{cases}\left|x-2018\right|\ge0\\\left|y-10\right|\ge0\\\left|z-1\right|\ge0\end{cases}}\text{dấu = xảy ra khi }\hept{\begin{cases}x=2018\\y=10\\z=1\end{cases}}\)

Vậy để \(\left|x-1000\right|+\left|x-2018\right|+\left|x-2019\right|+\left|y-10\right|+\left|z-1\right|=1019\) => \(\hept{\begin{cases}x=2018\\y=10\\z=1\end{cases}}\)

Tìm y  biết:

2020 - y : 8 x 10 = 1000

          y : 8 x 10 = 2000 - 1000

          y : 8 x 10 = 1000

          y : 8        = 1000 : 10

          y : 8        = 100

          y             = 100 x 8

          y             = 800

Vậy y = 800

\(2020-y:8\times10=1000\)

        \(2020-y:80=1000\)

                        \(y:80=2020-1000\)

                        \(y:80=1020\)

                              \(y=1020\times80\)

                              \(y=81600\)

2020-y:8=1000:10

2020-y:8=100

2020-y=100x8

2020-y=800

y=2020-800=1220

\(\dfrac{x}{5}=\dfrac{y}{2}=k\)\(\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)

\(\Rightarrow xy=10k^2=1000\Rightarrow k=\pm10\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=50\\y=20\end{matrix}\right.\\\left\{{}\begin{matrix}x=-50\\y=-20\end{matrix}\right.\end{matrix}\right.\)

 Ta có : y . 45 + y . 55 = 1000

            y ( 45 + 55 ) = 1000

            y . 100 = 1000

            y = 1000 : 100

            y = 10

 Vậy y = 10

y.45+y.55=1000

y.(45+55)=1000

y. 100      =1000

y               =1000:100

y              =10

Vậy y=10

y.45 + y.55 = 1000

y .(45 + 55) = 1000

y . 100 = 1000

         y = 1000 : 100

         y = 10