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

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

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

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 ..

( 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

a) \(y^{2015}=y^{2020}\)

\(\Leftrightarrow y^{2020}-y^{2015}=0\)

\(\Leftrightarrow y^{2015}.\left(y^5-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y^{2015}=0\\y^5-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}}\)

Vậy ...

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\left(2y-1\right)^1=0\\\left(2y-1\right)^{49}-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}y=\frac{1}{2}\\y=1\end{cases}}\)

Vậy...

c) \(\left(\frac{y}{3}-5\right)^{2020}=\left(\frac{y}{3}-5\right)^{2008}\)

\(\Leftrightarrow\left(\frac{y}{3}-5\right)^{2020}-\left(\frac{y}{3}-5\right)^{2008}\)

\(\Leftrightarrow\left(\frac{y}{3}-5\right)^{2008}.\left[\left(\frac{y}{3}-5\right)^{12}-1\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(\frac{y}{3}-5\right)^{2008}=0\\\left(\frac{y}{3}-5\right)^{12}-1=0\end{cases}}\)ra bên này có 3 TH nha vì TH cuối là mũ chẵn

a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+1+1}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

=>x+1=1 và y-2=1/2

=>x=0 và y=5/2

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{1}{2}-\dfrac{1}{18}=\dfrac{9}{18}-\dfrac{1}{18}=\dfrac{8}{18}=\dfrac{4}{9}\\\dfrac{2}{2x-y}=\dfrac{1}{18}+\dfrac{1}{x-2y}\end{matrix}\right.\)

=>x-2y=9 và 2/2x-y=1/18+1/9=1/18+2/18=3/18=1/6

=>x-2y=9 và 2x-y=12

=>x=5; y=-2

c: \(\Leftrightarrow\left\{{}\begin{matrix}10\left|x-6\right|+15\left|y+1\right|=25\\10\left|x-6\right|-8\left|y+1\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23\left|y+1\right|=23\\\left|x-6\right|=1\end{matrix}\right.\)

=>|x-6|=1 và |y+1|=1

=>\(\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)

x/2=y/3;y/2=z/5 => x/2=2y/6;3y/6=z/5 => x/4=y/6=z/15

adtcdtsbn:

x/4=y/6=z/15=x+y+z/4+6+15=50/25=2

suy ra : x/4=2=>x=4.2=8

y/6=2=>y=2.6=12

z/15=2 => z=15.2=30

Thì các bạn hãy trả lời từng câu 1

Bài 1 :

\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)

\(\Leftrightarrow x=-23\)

\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)