a: \(A=3-\left|3-x\right|\)

=3-|x-3|

TH1: x>=3

=>x-3>=0

A=3-|x-3|

=3-(x-3)

=3-x+3

=6-x

TH2: x<3

=>x-3<0

A=3-|x-3|

=3-(3-x)

=3-3+x

=x

b: \(B=\left|x-6\right|+\left|6-x\right|-2\)

=2|x-6|-2

TH1: x>=6

=>x-6>=0

B=2|x-6|-2

=2(x-6)-2

=2x-12-2

=2x-14

TH2: x<6

=>x-6<0

B=2|x-6|-2

=2(6-x)-2

=12-2x-2

=10-2x

c: \(C=\left|-x-1\right|+\left|-x-5\right|-x\)

\(=\left|x+5\right|+\left|x+1\right|-x\)

TH1: x<-5

=>x+5<0; x+1<0

\(C=\left|x+5\right|+\left|x+1\right|-x\)

=-x-5-x-1-x

=-3x-6

TH2: -5<=x<-1

=>x+5>=0; x+1<0

\(C=\left|x+5\right|+\left|x+1\right|-x\)

=x+5-x-1-x

=-x+4

TH3: x>=-1

=>x+5>0; x+1>=0

\(C=\left|x+5\right|+\left|x+1\right|-x\)

=x+5+x+1-x

=x+6

A=1+5²+5⁴+...+5²⁰⁰

5²A=5²+5⁴+...+5²⁰²

5²A+1=1+5²+5⁴+...+5²⁰⁰+5²⁰²=A+5²⁰²

25A-A=5²⁰²-1

24A=5²⁰²-1

A = \(\frac{5^{202}-1}{24}\)

Cảm ơn bạn nhiều nhé

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

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

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

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

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

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 ..

Vậy ta thấy 5A=5+5^2+5^3+5^4+...+5^2009+5^2010

=> 5A-A= 5^2010-1

=> 4A=5^2010-1=> 4A=(5^2010-1)/4

 đến đaay em tính ra bằng máy tính hay để nguyên thì chắc chắn cô giáo sẽ cho điểm, tốt nhất cứ để nguyên nhé :)

Nguyễn đức hiếu làm sai kìa 

Đoạn cuối :

4A = 5²⁰²⁰ -1 

\(A = { {5mũ2020-1} \over 4}\)