=> 2y-1 = 0 hoặc 2y-1=1

=> y = 1/2 hoặc y = 1

k mk nha

(2y-1)^50 = 2y-1

<=>(2y-1)^50-(2y-1) = 0

<=> (2y-1).[(2y-1)^49-1] = 0

=> 2y-1 = 0 hoặc (2y-1)^49 = 1 = 1^49

=> 2y-1 = 0 hoặc 2y-1 = 1

=> y=1/2 hoặc y=1

Để (2y - 1)^50 = 2y-1 thì suy ra 2y-1 = 1

Vậy y = 1 

Nhớ k nhé! Thank you!!!

là sao vậy aa

=> \(\left(2y-1\right)^{50}-\left(2y-1\right)=0\)

=>\(\left(2y-1\right)\left[\left(2y-1\right)^{49}-1\right]=0\)

=> 2y - 1 = 0 hoặc (2y-1)^49 - 1 =0

=> y = 1/2    hoặc (2y-1)^49 = 1

=> y=1/2      hoặc 2y-1=1  => y=1

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

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

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

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 ..

525km

(2x-1)⁵⁰=2x-1

(2x-1)⁵⁰=2x-1=0 hoặc= 1 (x thuộc Z)

   Nếu (2x-1)⁵⁰=2x-1=0

        2x-1=0

        x = 0,5 ( không thỏa mãn điều kiện )

 Nếu (2x-1)⁵⁰=2x-1=1

        2x-1=1

          x=1 ( thỏa mãn điêù kiện)

Vậy x=1

Y=0;1;-1

Y=0;1;-1

a) y^200 = y

\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=0\end{cases}}\)

b) y^2008 = y^2010

\(\Leftrightarrow\orbr{\begin{cases}y=1\\y=0\end{cases}}\)

c) (2y - 1)^50 = 2y - 1

\(\Leftrightarrow\orbr{\begin{cases}2y-1=1\\2y-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=1\\y=\frac{1}{2}\end{cases}}\)

d) (y/3 - 5)^2000= y/3 -5

\(\Leftrightarrow\orbr{\begin{cases}\frac{y}{3}-5=1\\\frac{y}{3}-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=18\\y=15\end{cases}}\)

a, y²⁰⁰=y

=>y²⁰⁰-y=0

=>y(y¹⁹⁹-1)=0

=>\(\orbr{\begin{cases}y=0\\y^{199}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}}\)

b, y²⁰⁰⁸=y²⁰¹⁰

=>y²⁰⁰⁸-y²⁰¹⁰=0

=>y²⁰⁰⁸(1-y²)=0

=>\(\orbr{\begin{cases}y^{2008}=0\\1-y^2=0\end{cases}\Rightarrow\orbr{\begin{cases}y=0\\y^2=1\end{cases}\Rightarrow}\orbr{\begin{cases}y=0\\y=\pm1\end{cases}}}\)

c,d tương tự a,b

( 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