1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) +...+ \(\dfrac{1}{x\left(x+1\right):2}\) = \(1\dfrac{1991}{1993}\)
Tìm x:
a) \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
b) \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=1\dfrac{1991}{1993}\)
a)
<=> (1/3)[3/(5.8) + 3/(8.11) + ... + 3/[x(x+3)] = 101/1540
<=> (1/3)[(1/5 - 1/8) + (1/8 - 1/11) + ... + 1/x - 1/(x+3)] = 101/1540
<=> (1/3)[1/5 - 1/(x+3)] = 101/1540
<=> 1/5 - 1/(x+3) = 303/1540
<=> 1/(x+3) = 1/5 - 303/1540 = 5/1540 = 1/308
<=> x = 305
b)
a)\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
\(\dfrac{1.3}{5.8}+\dfrac{1.3}{8.11}+\dfrac{1.3}{11.14}+...+\dfrac{1.3}{x.\left(x+3\right)}=\dfrac{101.3}{1540}\)
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}=\dfrac{303}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{308}\)
308.1 = (x + 3).1
308 = x + 3
x = 308 - 3
x = 305
a) \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
=> \(3.\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{x\left(x+3\right)}\right)=\dfrac{101}{1540}.3\)
=> \(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
=> \(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
=> \(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
=> \(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)
=> \(\dfrac{1}{x+3}=\dfrac{1}{308}\)
=> \(x+3=308\)
=> \(x=305\)
1. Tìm x biết:
a) \(\left(-x-\dfrac{1}{9}\right)^2=\dfrac{4}{9}\)
b) \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=1\dfrac{1991}{1993}\)
a)\(\left(-x-\dfrac{1}{9}\right)^2=\dfrac{4}{9}\)
\(\Rightarrow\left(-x-\dfrac{1}{9}\right)^2=\left(\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^2\)
*)Xét \(\left(-x-\dfrac{1}{9}\right)^2=\left(\dfrac{2}{3}\right)^2\)
\(\Rightarrow-x-\dfrac{1}{9}=\dfrac{2}{3}\Rightarrow-x=\dfrac{7}{9}\Rightarrow x=-\dfrac{7}{9}\)
*)Xét \(\left(-x-\dfrac{1}{9}\right)^2=\left(-\dfrac{2}{3}\right)^2\)
\(\Rightarrow-x-\dfrac{1}{9}=-\dfrac{2}{3}\Rightarrow-x=-\dfrac{5}{9}\Rightarrow x=\dfrac{5}{9}\)
b)\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}=1\dfrac{1991}{1993}\)
\(\Rightarrow\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}=\dfrac{1991}{1993}\)
\(\Rightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1991}{1993}\)
\(\Rightarrow\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1991}{1993}\)
\(\Rightarrow2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1991}{1993}\)
\(\Rightarrow\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1991}{3986}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1991}{3986}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1991}{3986}\)\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{1993}\)
\(\Rightarrow x+1=1993\Rightarrow x=1992\)
Tìm x biết:
\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.....+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{1999}{1993}\)
\(\Leftrightarrow\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=1+\dfrac{1999}{1993}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3992}{1993}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1996}{1993}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{1996}{1993}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{-3}{1993}\)
=>x+1=-1993/3
hay x=-1996/3
Tìm x, biết rằng:
a, \(\dfrac{1}{5\cdot8}\) + \(\dfrac{1}{8\cdot11}\) + \(\dfrac{1}{11\cdot14}\) + ... + \(\dfrac{1}{x\left(x+3\right)}\) = \(\dfrac{101}{1540}\)
b, 1+\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{x\left(x+1\right)\div2}\) = \(1\dfrac{1991}{1993}\)
Các bạn giúp tớ với, sáng mai mình học rồi.
a)\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
a) \(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+\(\dfrac{1}{11.14}\)+...+\(\dfrac{1}{x\left(x+3\right)}\)=\(\dfrac{101}{1540}\)
=\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)
=\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
=> \(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
=> \(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)
=> \(\dfrac{1}{x+3}=\dfrac{1}{308}\)
=> x+3=308
=> x=308-3
=> x=305
b) \(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{1991}{1993}\)
= \(\dfrac{1}{2}.\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}\right)=\dfrac{3984}{3986}\)
= \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{3984}{3986}\)
= \(2-\dfrac{1}{1.2}+3-\dfrac{2}{2.3}+4-\dfrac{3}{3.4}+...+x+1-\dfrac{x}{x\left(x+1\right)}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
= \(1-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
=> \(\dfrac{1}{x+1}=1-\dfrac{3984}{3986}\)
=> \(\dfrac{1}{x+1}=\dfrac{2}{3986}=\dfrac{1}{1993}\)
=> x+1= 1993
=> x= 1993-1
=> x=1992
1 .Tìm x biết
a. ( \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\)) = \(\dfrac{0,33x}{2009}\)
b. 1 + \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1991}{1993}\)
c. \(\dfrac{1}{2013}x+1+\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{2012.2013}=2\)
d. 2x + \(\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}=10\)
2. Chứng minh rằng :
a. \(\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{99}}< \dfrac{1}{3}\)
b. \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
1/
a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
Bài 1:
c: \(\Leftrightarrow x\cdot\dfrac{1}{2013}+1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}-\dfrac{1}{2013}=2\)
\(\Leftrightarrow x\cdot\dfrac{1}{2013}+2-\dfrac{1}{2013}=2\)
=>x*1/2013=1/2013
=>x=1
d: \(\Leftrightarrow2x+8+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}=10\)
=>2x+1/2-1/3+1/3-1/4+...+1/9-1/10=2
=>2x+2/5=2
=>x+1/5=1
=>x=4/5
Bài 3: Tìm x:
a) \(\dfrac{1}{5.8}\)+\(\dfrac{1}{8.11}\)+\(\dfrac{1}{11.14}\)+......+\(\dfrac{1}{x\left(x+3\right)}\)= \(\dfrac{101}{1540}\)
b) 1+ \(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+......+ \(\dfrac{1}{x\left(x+1\right):2}\)=1\(\dfrac{1991}{1993}\)
Ai biết làm bài này thì giúp mik nhé!
mik đang cần gấp
Cảm ơn nhiều!♥
\(a,\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.3.\left[\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left[\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left[\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right]=\dfrac{101}{1540}\)
\(\dfrac{1}{3}.\left(\dfrac{1}{5-1}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}.\dfrac{1}{3}\)
\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{3}-\dfrac{303}{1540}\)
\(\dfrac{1}{x+3}=\dfrac{1}{308}\)
\(\Rightarrow x+3=308\)
\(x=308-3\)
\(x=305\)
\(b,1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=1\dfrac{1991}{1993}\)
\(\dfrac{1}{2}.\left(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x.\left(x+1\right):2}\right)=\dfrac{3984}{3986}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{8}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+x+1-\dfrac{x}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
\(1-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)
\(\dfrac{1}{x+1}=1-\dfrac{3984}{3986}\)
\(\dfrac{1}{x+1}=\dfrac{1}{1993}\)
=>\(x+1=1993\)
\(x=1993-1\)
\(x=1992\)
1. \(\dfrac{5\left(x-1\right)+2}{6}-\dfrac{7x-1}{4}=\dfrac{2\left(2x+1\right)}{7}-5\)
2. \(x-\dfrac{3\left(x+30\right)}{15}-24\dfrac{1}{2}=\dfrac{7x}{10}-\dfrac{2\left(10x+2\right)}{5}\)
3. \(14\dfrac{1}{2}-\dfrac{2\left(x+3\right)}{5}=\dfrac{3x}{2}-\dfrac{2\left(x-7\right)}{3}\)
4. \(\dfrac{x+1}{3}+\dfrac{3\left(2x+1\right)}{4}=\dfrac{2x+3\left(x+1\right)}{6}+\dfrac{7+12x}{12}\)
5. \(\dfrac{3\left(2x-1\right)}{4}-\dfrac{3x+1}{10}+1=\dfrac{2\left(3x+2\right)}{5}\)
6. \(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)
7. \(\dfrac{3\left(x-3\right)}{4}+\dfrac{4x-10,5}{10}=\dfrac{3\left(x+1\right)}{5}+6\)
8. \(\dfrac{2\left(3x+1\right)+1}{4}-5=\dfrac{2\left(3x-1\right)}{5}-\dfrac{3x+2}{10}\)
1,\(\dfrac{5\left(x-1\right)+2}{6}\)-\(\dfrac{7x-1}{4x}\)=\(\dfrac{2\left(2x+1\right)}{7}\)-5
2,\(\dfrac{3\left(x-3\right)}{4}\)+\(\dfrac{4x-10,5}{10}\)=\(\dfrac{3 \left(x+1\right)}{5}\)+6
3,\(\dfrac{2\left(3x+1\right)+1}{4}\)-5=\(\dfrac{2\left(3x-1\right)}{5}\)-\(\dfrac{3x+2}{10}\)
Diễn giải ra cho em với ạ!Em cảm ơn
1, bạn xem lại đề
2, 15(x-3) + 8x-21 = 12(x+1) +120
<=> 23x - 66 = 12x + 132
<=> 11x = 198 <=> x = 198/11
3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4
<=> 30x + 10 - 95 = 18x -12
<=> 12x = 73 <=> x = 73/12
Tìm x biết: a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\) b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\) d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}.\dfrac{10}{6}\)
a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)
Th1 : \(x-\dfrac{1}{2}=0\)
\(x=0+\dfrac{1}{2}\)
\(x=\dfrac{1}{2}\)
Th2 : \(-3-\dfrac{x}{2}=0\)
\(\dfrac{x}{2}=-3\)
\(x=\left(-3\right)\cdot2\)
\(x=-6\)
Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)
b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)
\(x=\dfrac{5}{8}+\dfrac{1}{8}\)
\(x=\dfrac{3}{4}\)
c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)
\(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)
\(\dfrac{3}{2}+x=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{2}\)
\(x=0\)
d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)
\(x+\dfrac{1}{3}=-4\)
\(x=-4-\dfrac{1}{3}\)
\(x=-\dfrac{13}{3}\)