Các số sau theo thứ tự từ bé đến lớn: 499, 512, 521, 1000

\(\frac{1}{3}+\frac{1}{6}=\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{499}{1000}\)

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{499}{1000}\)

\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{499}{1000}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{499}{1000}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{499}{1000}\)

\(2\left(\frac{1}{2}-\frac{1}{x-1}\right)=\frac{499}{1000}\)

\(\frac{1}{2}-\frac{1}{x-1}=\frac{499}{2000}\)

\(\frac{2000\left(x-1\right)}{2\left(x-1\right)2000}-\frac{2.2000}{\left(x-1\right)2.2000}=\frac{499.2\left(x-1\right)}{2000.2\left(x-1\right)}\)

Khử mẫu 

\(2000x-2000-4000=998x-998\)

\(2000x-6000=998x-998\)

\(1002x-5002=0\)

\(1002x=5002\Leftrightarrow x=\frac{2501}{501}\)

theo bai co 1/3+1/6+...+2/x.x+1=499/1000

(=)2/2.3+2/3.4+...+2/x.(x+1)=499/1000

(=)2(1/2-...-1/x+1)=499/1000

(=)1/2-1/x+1=499/2000

(=)1/x+1=501/2000

den day thi minh chiu roi!

69 245 < 69 260

73 500 > 73 499

60 000 = 59 000 + 1000

70 000 + 30 000 = 100 000

20 000 + 40 000 < 60 600

80 000 + 8000 > 80 900

cm cái j?

cm 1<A*A<4

cm < 1 à?

Mình giải luôn ra nha

Tìm x:
1/3+1/6+1/10+.........+2/x.(x+1)=499/1000
1/2.(1/3+1/6+1/10+.......+2/x.(x+1)=499/1000.1/2
1/6+1/12+1/20+.......+1/x.(x+1)=499/2000
1/(2.3)+1/(3.4)+1/(4.5)+.........+1/x.(x+1)=499/2000
1/2-1/3+1/3-1/4+1/4-1/5+........+1/x-1/(x+1)=499/2000
1/2-1/(x+1)=499/2000
1/(x+1)=1/2-499/2000
1/(x+1)=501/2000
\Rightarrow1.2000=(x+1).501
\Rightarrow2000=x.501+501
\Rightarrow1499=x.501
\Rightarrowx=1499:501
Vì x thuộc Z nên 1499:501 là 1 số nguyên.Mà 1499:501 được 1 số thập phân nên x thuộc rỗng.