\(\Leftrightarrow\dfrac{2032-x}{25}-1+\dfrac{2053-x}{23}-2+\dfrac{2070-x}{21}-3+\dfrac{2083-x}{19}-4=0\)

=>2007-x=0

hay x=2007

\(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}=10\)

\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)

\(\Leftrightarrow\dfrac{2032-x-25}{25}+\dfrac{2053-x-46}{23}+\dfrac{2070-x-63}{21}+\dfrac{2083-x-76}{19}=0\)

\(\Leftrightarrow\dfrac{2007-x}{25}+\dfrac{2007-x}{23}+\dfrac{2007-x}{21}+\dfrac{2007-x}{19}=0\)

\(\Leftrightarrow\left(2007-x\right)\left(\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\right)=0\)

\(\Leftrightarrow2007-x=0\left(vì.\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}+\dfrac{1}{19}\ne0\right)\)

\(\Leftrightarrow x=2007\)

\(2x+\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}=10\)

\(2x+\left(1+\dfrac{1}{6}\right)+\left(1+\dfrac{1}{12}\right)+\left(1+\dfrac{1}{20}\right)+\left(1+\dfrac{1}{30}\right)+\left(1+\dfrac{1}{42}\right)+\left(1+\dfrac{1}{56}\right)+\left(1+\dfrac{1}{72}\right)+\left(1+\dfrac{1}{90}\right)=10\)

\(2x+8+\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=10\)

\(2x+8+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)=10\)

\(2x+8+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)=10\)

\(2x+8+\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=10\)

~ Chúc bạn học tốt ~

2x+\(\dfrac{7}{2.3}+\dfrac{13}{3.4}+\dfrac{21}{4.5}+...+\dfrac{91}{9.10}=10\)

\(2x+7\left(\dfrac{1}{2.3}\right)+13\left(\dfrac{1}{3.4}\right)+...+91\left(\dfrac{1}{9.10}\right)=10\)

\(2x+\left(7+13+...+91\right)\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right)=10\)

\(2x+336+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=10\)

\(2x+336+\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=10\)

\(2x+336.\dfrac{2}{5}=10\)

\(2x+\dfrac{672}{5}=10\)

\(2x=10-\dfrac{672}{5}\)

\(2x=-\dfrac{622}{5}\)

\(x=-\dfrac{311}{5}\)

Tick nha 0o0^^^Nhi^^^0o0

d: ĐKXĐ: x<>-4; x<>-5; x<>-6; x<>-7

\(PT\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

=>x^2+11x+28=54

=>x^2+11x-26=0

=>(x+13)(x-2)=0

=>x=2 hoặc x=-13

e: \(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)

\(\Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\)

=>x-258=0

=>x=258

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

????????????????

mik còn chx hc nx

=\(\dfrac{35}{36}.\dfrac{-36}{35}< x< \dfrac{13}{8}.\dfrac{8}{13}\)

 = -1 < x < 1

vậy x = 0

D

chịu

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)

\(\dfrac{1}{3}\)  - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)

 \(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) =  \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) +  \(\dfrac{1}{11}\) =  \(x\) - \(\dfrac{5}{13}\)

         \(x-\dfrac{5}{13}=\dfrac{1}{11}\)

        \(x\)           = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)

      \(x\)           = \(\dfrac{68}{143}\)

Em cảm ơn ạ.

Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}\)\(=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1980}{19}\)

\(\Leftrightarrow\left(\frac{x-29}{1970}-1\right)+\left(\frac{x-27}{1972}-1\right)+\left(\frac{x-25}{1974}-1\right)+\left(\frac{x-23}{1976}-1\right)+\left(\frac{x-21}{1978}-1\right)+\left(\frac{x-19}{1980}-1\right)\)\(=\left(\frac{x-1970}{29}-1\right)+\left(\frac{x-1972}{27}-1\right)+\left(\frac{x-1974}{25}-1\right)+\left(\frac{x-1976}{23}-1\right)+\left(\frac{x-1978}{21}-1\right)+\left(\frac{x-1980}{19}-1\right)\)

\(\Leftrightarrow\frac{x-1999}{1970}+\frac{x-1999}{1972}+\frac{x-1999}{1974}+\frac{x-1999}{1976}+\frac{x-1999}{1978}+\frac{x-1999}{1980}\)\(=\frac{x-1999}{29}+\frac{x-1999}{27}+\frac{x-1999}{25}+\frac{x-1999}{24}+\frac{x-1999}{21}+\frac{x-1999}{19}\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}\right)\)\(=\left(x-1999\right)\left(\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}-\frac{1}{23}-\frac{1}{21}-\frac{1}{19}\right)=0\)\(\Leftrightarrow\) \(x-1999=0\) (Vì ...khác 0)

\(\Leftrightarrow x=1999\)(thỏa mãn)

Vậy \(x=1999\)