a).  x^4  =  x^7

=>  x   =  \(\orbr{\begin{cases}0\\1\end{cases}}\)          

b)  (2y+1)^6 = (2y+1)^8

=>    \(\orbr{\begin{cases}2y+1=0\\2y+1=1\end{cases}}\)       

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

`a,`

`x^4=x^7`

`->x^4-x^7=0`

`->x^4 (1 - x^3)=0`

TH1 : `x^4=0 ->x=0`

TH2 : `1-x^3=0 ->x^3=1^3 ->x=1`

Vậy `x=0,x=1`

`b,`

`(2y+1)^6 = (2y+1)^8`

`-> (2y+1)^6 - (2y+1)^8=0`

`-> (2y+1)^6 [1-(2y+1)^2]=0`

TH1 : `(2y+1)^6 =0 ->2y+1=0 ->y=(-1)/2`

TH2 : `1 - (2y+1)^2=0`

`-> (2y+1)^2=1`

`-> 2y+1=1` hoặc `2y+1=-1`

`->y=0` hoặc `y=-1`

Vậy `y=0,y=-1,y=(-1)/2`

Trả lời câu hỏi chính dùm 

Vào câu hỏi của ng khác xong hỏi là zô zuyên nhé 

Ht 

____ Vanilla ___

\(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{1}{2008}\)

\(=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)\)

\(=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2008}=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\)

Khi đó \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}{2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)}=\frac{1}{2009}\)

gạch chéo là chia ạ

Giải :

\(\left|-3,45\right|+\frac{1,5}{3}\)

\(=3,45+0,5\)

\(=3,95\)

Đấy nha

thanks

2 mu 4 -x=32

16-x=32

    x=16-32

   x=-16

\(\frac{a}{b}-\frac{a+2001}{b+2001}=\frac{a\left(b+2001\right)-b\left(a+2001\right)}{b\left(b+2001\right)}=\frac{2001\left(a-b\right)}{b\left(b+2001\right)}.\)

Ta có \(b>0\Rightarrow b\left(b+2001\right)>0\)

+ Nếu \(a>b\Rightarrow2001\left(a-b\right)>0\Rightarrow\frac{2001\left(a-b\right)}{b\left(b+2001\right)}>0\Rightarrow\frac{a}{b}>\frac{a+2001}{b+2001}\)

+ Nếu \(a< b\Rightarrow2001\left(a-b\right)< 0\Rightarrow\frac{2001\left(a-b\right)}{b\left(b+2001\right)}< 0\Rightarrow\frac{a}{b}< \frac{a+2001}{b+2001}\)

X mũ 2020+x = 0

 x=0

x\(^{2020}\)+ x = x\(^{2021}\) 

Để \(A\inℤ\Rightarrow x+1⋮3\)

Đặt x + 1 = 3k (k \(\inℤ\)) 

=> x = 3k - 1

mà \(x\inℤ\)

Vậy \(x=3k-1\left(k\inℤ\right)\)