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!

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.

\(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^x+6^{x+2}=6^{10}+6^7\)

\(\frac{1}{6}\cdot6^x+6^{x+2}=6^7\cdot\left(6^3+1\right)\)

\(6^{x-1}\cdot\left(6^3+1\right)=6^7\cdot\left(6^3+1\right)\)

\(x-1=7\)

\(x=8\)

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

\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)

mik lỡ bấm nhầm rồi, phần sau bn tự nghĩ nhé, sorry

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

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

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

2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-..........+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\))=\(\dfrac{499}{1000}\)

2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\))=\(\dfrac{499}{1000}\)

\(\dfrac{1}{2}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{499}{1000}\):2

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

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

\(\dfrac{1}{x+1}\)=\(\dfrac{501}{2000}\)

Đề bài sai phải

(x+2).(y-3)=-3=-1.3=1.(-3)

Vì x,y thuộc Z nên ( x+2) và (y+3) thuộc Z

Ta có bảng:

Vậy nếu x = - 3 thì y = 0

       nếu x = -1 thì y =- 6

       nếu x = - 5 thì y = - 2

       nếu x = 1 thì y = - 4

sao ko làm hết  bài

a) 1 - 2 + 3 - 4 + 5 - 6 + ... + 159 - 160

= (1 - 2) + (3 - 4) + (5 - 6) + ... + (159 - 160)

= (-1) + (-1) + (-1) + ... + (-1)

       có 80 số -1

= (-1) . 80

= -80

1 - 2 + 3 - 4 + 5 - 6 + ......... + 159 - 160 ( có 160 số )

= - 1 + ( - 1 ) + ( - 1 ) + .......... + ( - 1 )    ( có 80 số - 1 )

= - 1 . 80

= - 80 

Bài 2 :

a ) l x l < 3

=> l x l thuộc { 0 ; 1 ; 2 }

=> x thuộc { - 2 ; - 1 ; 0 ; 1 ; 2 }

Vậy x thuộc  { - 2 ; - 1 ; 0 ; 1 ; 2 }