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a)\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)

\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)

\(a,\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)

\(\dfrac{1}{3}.3.\left[\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)

\(\dfrac{1}{3}.\left[\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right]=\dfrac{101}{1540}\)

\(\dfrac{1}{3}.\left[\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right]=\dfrac{101}{1540}\)

\(\dfrac{1}{3}.\left(\dfrac{1}{5-1}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}.\dfrac{1}{3}\)

\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\dfrac{1}{x+3}=\dfrac{1}{3}-\dfrac{303}{1540}\)

\(\dfrac{1}{x+3}=\dfrac{1}{308}\)

\(\Rightarrow x+3=308\)

\(x=308-3\)

\(x=305\)

\(b,1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(x+1\right):2}=1\dfrac{1991}{1993}\)

\(\dfrac{1}{2}.\left(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x.\left(x+1\right):2}\right)=\dfrac{3984}{3986}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{8}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)

\(\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+x+1-\dfrac{x}{x.\left(x+1\right)}=\dfrac{3984}{3986}\)

\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}+\dfrac{1}{x+1}=\dfrac{3984}{3986}\)

\(1-\dfrac{1}{x+1}=\dfrac{3984}{3986}\)

\(\dfrac{1}{x+1}=1-\dfrac{3984}{3986}\)

\(\dfrac{1}{x+1}=\dfrac{1}{1993}\)

=>\(x+1=1993\)

\(x=1993-1\)

\(x=1992\)

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)\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}=1\dfrac{1991}{1993}\)

\(\Rightarrow\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{\dfrac{x\left(x+1\right)}{2}}=\dfrac{1991}{1993}\)

\(\Rightarrow\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1991}{1993}\)

\(\Rightarrow\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+\dfrac{2}{4\cdot5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1991}{1993}\)

\(\Rightarrow2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{1991}{1993}\)

\(\Rightarrow\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1991}{3986}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1991}{3986}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1991}{3986}\)\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{1993}\)

\(\Rightarrow x+1=1993\Rightarrow x=1992\)

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) \(\left(2x-3\right)\left(6-2x\right)=0\)

\(\circledast\)TH₁: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)

\(\circledast\)TH₂: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)

Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).

b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)

\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)

\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)

\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)

\(-\dfrac{11}{15}=-x\left(x-1\right)\)

\(\Rightarrow x=1.491631652\)

Vậy \(x=1.491631652\)

c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)

\(\circledast\)TH₁: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)

\(\circledast\)TH₂: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)

Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).

d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)

Vậy \(x=\dfrac{10}{3}\).

e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)

\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)

\(\dfrac{x}{3}=\dfrac{7}{10}\)

\(x=\dfrac{3\cdot7}{10}\)

\(x=\dfrac{21}{10}\)

Vậy \(x=\dfrac{21}{10}\).

f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)

\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)

\(\dfrac{x}{5}=\dfrac{11}{10}\)

\(x=\dfrac{5\cdot11}{10}\)

\(x=\dfrac{55}{10}=\dfrac{11}{2}\)

Vậy \(x=\dfrac{11}{2}\).

g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)

Vậy \(x=2\).

h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)

Vậy \(x=14\).

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

\(B=\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\left(1+\dfrac{1}{24}\right).....\left(1+\dfrac{1}{440}\right)\left(1+\dfrac{1}{483}\right)\)

\(B=\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}.....\dfrac{441}{440}.\dfrac{484}{483}\)

\(B=\dfrac{9.16.25.....441.484}{8.15.24.....440.483}\)

\(B=\dfrac{3.3.4.4.5.5.....21.21.22.22}{2.4.3.5.4.6.....20.22.21.23}\)

\(B=\dfrac{3.4.5.....21.22}{2.3.4.....20.21}.\dfrac{3.4.5.....21.22}{4.5.6.....22.23}\)

\(B=11.\dfrac{3}{23}=\dfrac{33}{23}\)

B = \(\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{25}{24}...\dfrac{121}{120}.\dfrac{144}{143}\)

B = \(\dfrac{4.9.16.25...121.144}{3.8.15.24....120.143}\)

B = \(\dfrac{2.2.3.3.4.4.5.5...11.11.12.12}{1.3.2.4.3.5.4.6...10.12.11.13}\)

B = \(\dfrac{2.3.4.5...11.12}{1.2.3.4.5...10.11}.\dfrac{2.3.4.5...11.12}{3.4.5.6.7...12.13}\)

B = 12 . \(\dfrac{2}{13}\)

B = \(\dfrac{24}{13}\)

\(1+\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+2\right)}=1\dfrac{2009}{2011}\)

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{4020}{4022}\)

\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{4020}{4022}\)

\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4020}{4022}\)

\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{4020}{4022}\)

\(\Rightarrow\dfrac{1}{x+2}=\dfrac{1}{2011}\)

\(\Rightarrow x+2=2011\Rightarrow x=2009\)

Vậy x = 2009

a) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)

\(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\)

\(-\dfrac{2}{3}x=\dfrac{1}{10}\)

x=\(\dfrac{1}{10}:-\dfrac{2}{3}\)

\(x=-\dfrac{3}{20}\)

Vậy \(x=-\dfrac{3}{20}\).

b) \(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)

\(\dfrac{2}{3}:x=-7-\dfrac{1}{3}\)

\(\dfrac{2}{3}:x=-\dfrac{22}{3}\)

\(x=\dfrac{2}{3}:-\dfrac{22}{3}\)

\(x=-\dfrac{1}{11}\)

Vậy \(x=-\dfrac{1}{11}\).

c) \(60\%x=\dfrac{1}{3}\cdot6\dfrac{1}{3}\)

\(60\%x=\dfrac{19}{9}\)

\(\dfrac{3}{5}x=\dfrac{19}{9}\)

\(x=\dfrac{19}{9}:\dfrac{3}{5}\)

\(x=\dfrac{95}{27}\)

Vậy \(x=\dfrac{95}{27}\).

d) \(\left(\dfrac{2}{3}-x\right):\dfrac{3}{4}=\dfrac{1}{5}\)

\(\dfrac{2}{3}-x=\dfrac{1}{5}\cdot\dfrac{3}{4}\)

\(\dfrac{2}{3}-x=\dfrac{3}{20}\)

\(x=\dfrac{2}{3}-\dfrac{3}{20}\)

\(x=\dfrac{31}{60}\)

Vậy \(x=\dfrac{31}{60}\).

e) \(-2x-\dfrac{-3}{5}:\left(-0.5\right)^2=-1\dfrac{1}{4}\)

\(-2x-\dfrac{-12}{5}=-1\dfrac{1}{4}\)

\(-2x=-1\dfrac{1}{4}+\dfrac{-12}{5}\)

\(-2x=-\dfrac{73}{20}\)

\(x=-\dfrac{73}{20}:\left(-2\right)\)

\(x=\dfrac{73}{40}\)

Vậy \(x=\dfrac{73}{40}\).

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