Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{98}{1545}\)
<=>\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{98}{1545}\)
<=>\(\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{x+3}\right)=\frac{98}{1545}\)
<=>\(\frac{1}{3}-\frac{1}{x+3}=\frac{98}{1545}:\frac{1}{3}\)
<=>\(\frac{1}{3}-\frac{1}{x+3}=\frac{98}{515}\)
<=>\(\frac{1}{x+3}=\frac{1}{3}-\frac{98}{515}\)
<=>\(\frac{1}{x+3}=\frac{221}{1545}\)
<=> \(x=?????\)
Hình như đề sai hay sao vậy bạn?
hai dòng dưới đề ý trong ngoặc lúc đầu đâu có \(\frac{1}{3}\)
còn chi tiết đây
a)1/5.8+1/8.11+1/11.14+...+1/x.(x+3)=101/1540
1/(5.8)+1/(8.11)+1/(11.14)+...1/x.(… =101/1540
3/(5.8)+3/(8.11)+...+3/x(x+3)=3.(10…
1/5-1/8+1/8-1/11+...+1/x-1/(x+3)=30…
1/5-1/(x+3)=303/1540
1/(x+3)=1/5-303/1540=1/308
=>x=305
lời giải nè : ấn vô dòng đen đen ở dưới ấy nhé
Tìm x, biết:a) 1/5.8 + 1/8.11 + 1/11.14 + ... + 1/x.(x+3)= 101/1540b) 1+ 1/3 + 1/6 + 1/10 +...+ 1/x.(x+1):2 = $1\frac{1991}{1993}$119911993
\(a)\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x(x+3)}=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left[(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3})\right]=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{3}\left[\frac{1}{5}-\frac{1}{x+3}\right]=\frac{101}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Rightarrow x=305\)
\(b)x-(\frac{50x}{100}-\frac{25x}{200})=\frac{45}{4}\)
\(\Rightarrow x-(\frac{100x}{200}-\frac{25x}{200})=\frac{45}{4}\)
\(\Rightarrow x-\frac{5x}{8}=\frac{45}{4}\)
\(\Rightarrow\frac{3x}{8}=\frac{45}{4}\)
\(\Rightarrow3x=\frac{45}{4}\cdot8\)
\(\Rightarrow3x=90\Rightarrow x=30\)
\(c)1+2+3+4+...+x=820\)
Ta có : \(1+2+3+4+...+x=\frac{(1+x)\cdot x}{2}\)
Do đó : \(\frac{(1+x)\cdot x}{2}=820\)
\(\Rightarrow(1+x)\cdot x=820\cdot2\)
\(\Rightarrow(1+x)\cdot x=1640\)
\(\Rightarrow(1+x)\cdot x=40\cdot41\)
Vì x và x + 1 là hai số tự nhiên liên tiếp nên => x = 40
Chúc bạn học tốt :3
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{1}{3}\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{x-2}{5x+15}=\dfrac{303}{1540}\)
\(\Rightarrow1540x-3080=1515x+4545\)
\(\Rightarrow25x=7625\)
\(\Rightarrow x=305\)
Vậy x = 305
Ta có:
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x\left(x+3\right)}=3.\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}=\dfrac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
Vậy \(x=305\)
=> 3B = 3.( 1/2.5 + 1/5.8 + 1/8.11 + ........... + 1/122.125)
= 3/2.5 + 3/5.8 + 3/ 8.11 + ......+ 3/122.125
Ta có: 3/ 2.5 = 1/2 - 1/5
3/5.8 = 1/5 -1/8
3/ 8.11 = 1/8 -1/11
..........................
3/122 . 125 = 3/122 - 3/125
=> 3B= 1/2 - 15/5 + 1/5 -1/8 +1/8 - 1/11 +........+1/122 - 1/125
= 1/2 - 1/125 = 125/250 - 2/250= 123/250
=> B= 3B : 3 = 123/250 :3 = 123/250 . 1/3 = 41/250
=> 2C = 2.(1/9.11 + 1/11.13 +....+ 1/97 .99)
= 2/9.11 + 2/11 .13 +.....+ 2/ 97.99
Ta có: 2/9.11 = 1/9 - 1/11
2/11.13 = 2/11 -2/ 13
...............................
2/97.99 = 1/97 - 1/99
=> 2B = 1/9 - 1/11 + 1/11 - 1/13 + ....+ 1/97 - 1/99
= 1/9 -1/99 = 11/99 - 1/99 =10/99
=> B= 2B : B = 10/99 :2 =10/99 . 1/2 = 5/99
Vậy B = 5/99
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)
⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)
Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.
Đặt 1/5.8 + 1/8.11 +...+ 1 /x (x+3) = A
3A = 3/5.8 + 3/8.11 +...+ 3/x (x+3)
3A = 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/x - 1/x+3
3A = 1/5 - 1/x + 3
3A = ( 3+x)-5/5x +15
A =[ ( 3+ x ) - 5 / 5x + 15 ] : 3
A = x + ( - 2 ) / 5x + 15
Ta có :
A + 27/480
= x + ( - 2 ) / 5x + 15
=> x + ( - 2 ) = 27
=> 5x + 15 = 480
* Làm nốt *
#Louis
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{27}{480}\)
\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{27}{480}\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{27}{480}\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{27}{480}\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{27}{480}.3\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{81}{480}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{81}{480}=\frac{15}{480}=\frac{1}{32}\)
\(\Rightarrow x+3=32\)
\(\Rightarrow x=32-3=29\)
ôn lại quy luât đi bài này dễ