![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/3+1/6+1/10+...+1/x*(2x+1)=1999/2001
2/6+2/12+...2/x(x+1)=1999/2001
2[1/2*3+1/3*4+...+1/x(x+1)]=1999/2001
1/2-1/3+1/3-1/4+...+1/x-1/x+1=1999/2001:2
(1/2-1/x+1)+(1/3-1/3)+...+(1/x-1/x)=1999/4002
1/2-1/x+1=1999/4002
1/x+1=1/2-1999/4002
1/x+1=1/2001
=>(x+1)=2001
x=2001-1
x=2000
Vậy x=2000
![](https://rs.olm.vn/images/avt/0.png?1311)
(*) <=> 1\6 + 1\12 +.. + 1\x.(x+1) = 2009\(2011.2)
ma
1\2.3 =1\2-1\3
1\3.4=1\3-1\4
...............
1\x(x+1)= 1\x-1\(x+1)
cong tung ve ta dc
Vt= 1\2- 1\(x+1) =2009\(2.2011)
<=> 2011\(2.2011) -2009\(2.2011) =1\(x+1)
<=> 1\2011 =1\(x+1)
=> x=2010
1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 1999/2001
nhân 1/2 vào 2 vế ta được vế trái là :
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{x-1}{2.\left(x+1\right)}=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{x-1}{\left(x+1\right)}=\frac{1999}{2001}\)
suy ra : 2001x - 2001 = 1999x + 1999
2x = 1999 + 2001 = 4000
=> x = 2000
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{-15}{18}-\left(x-\frac{1}{3}\right)=\frac{25}{27}\)
\(\frac{-5}{6}-x+\frac{2}{6}=\frac{25}{27}\)
\(\frac{-1}{2}-x=\frac{25}{27}\)
\(x=\frac{-77}{54}\)
Vậy............
b) \(\frac{-3}{5}-\left(2x-\frac{1}{20}\right)=\frac{3}{4}\)
\(\frac{-12}{20}-2x+\frac{1}{20}=\frac{15}{20}\)
\(\frac{-11}{20}-2x=\frac{15}{20}\)
\(2x=\frac{-13}{10}\)
\(x=\frac{-13}{20}\)
Vậy.............
1.
\(a,-\frac{15}{18}-\left(x-\frac{1}{3}\right)=\frac{25}{27}\)
\(-\frac{5}{6}-x+\frac{2}{6}=\frac{25}{27}\)
\(-\frac{1}{2}-x=\frac{25}{27}\)
\(x=-\frac{77}{54}\)
\(b,-\frac{3}{5}-\left(2x-\frac{1}{20}\right)=\frac{3}{4}\)
\(-\frac{12}{20}-2x+\frac{1}{20}=\frac{15}{20}\)
\(-\frac{11}{20}-2x=\frac{15}{20}\)
\(2x=-\frac{13}{10}\)
\(x=-\frac{13}{20}\)
2.
\(a,-\frac{5}{6}\)và \(1,2\)
\(=-\frac{5}{6}\)và \(\frac{12}{10}\)
\(=-\frac{50}{60}\)và \(\frac{72}{60}\)
Nếu như quy đồng 2 số lên thì ta đc \(-\frac{50}{60}< \frac{72}{60}\)
\(\Rightarrow-\frac{5}{6}\)\(< 1,2\)
\(b,\frac{15}{16}\)và \(\frac{17}{18}\)
Theo như những bài toán đã hc thìn ội dung ở cuối bài là phân số nào có tử bé hơn thì có phân số lớn hơn phân số có tử lớn hơn
\(\Rightarrow\frac{15}{16}>\frac{17}{18}\)
\(c,\frac{1999}{2000}\)và \(\frac{2000}{2001}\)
Ta quy đồng
Đc
\(\frac{3999999}{4002000}\)và \(\frac{4000000}{4002000}\)
\(\Rightarrow\frac{1999}{2000}< \frac{2000}{2001}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
quy dong TS tat ca len 2
2/6+2/12+2/20+...+2/x(x+1)
=2/2.3+2/3.4+2/4.5+...+2/x.(x+1)
=1/2-1/3+1/3-1/4+1/4-1/5+....+1/x-1/x+1
=1/2-1/x+1=1999/2001
![](https://rs.olm.vn/images/avt/0.png?1311)
1/3 + 1/6 + 1/10 + ... + 2/x(x + 1) = 1999/2001
2 × (1/6 + 1/12 + 1/20 + ... + 1/x(x + 1) = 1999/2001
1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x + 1) = 1999/2001 : 2
1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 1999/2001 × 1/2
1/2 - 1/x+1 = 1999/4002
1/x+1 = 1/2 - 1999/4002
1/x+1 = 2/4002 = 1/2001
=> x + 1 = 2001
=> x = 2001 - 1 = 2000
Vậy x = 2000
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - ...... + 1998 - 1999 - 2000 + 2001 + 2002
S = 1 + (2 - 3 - 4 + 5 )+ (6 - 7 - 8 + 9) + (10 - ...... + (1998 - 1999 - 2000 + 2001) + 2002
S=1+0+0...+0+2002
S= 1+2002
S=2003
Lời giải:
$S=(1+2-3-4)+(5+6-7-8)+(9+10-11-12)+...+(1997+1998-1999-2000)+2001+2002$
$=\underbrace{(-4)+(-4)+....+(-4)}_{500}+2001+2002$
$=(-4).500+2001+2002=2003$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
mik nghĩ chỗ \(\dfrac{2}{x.\left(x+1\right)}\) phải là \(\dfrac{1}{x.\left(x+1\right)}\) bạn có thể vui lòng kiểm tra lại đề không Lệ Quyên
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2001}{2003}\)
\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2001}{2003}\)
\(\Leftrightarrow\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2001}{2003}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2001}{4006}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2001}{4006}\)\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2003}\)
\(\Leftrightarrow x+1=2003\Leftrightarrow x=2002\)