a: \(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}-10=0\)

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

=>2007-x=0

hay x=2007

b: \(\Leftrightarrow x+\left(1+1+1+1+1+1+1\right)+\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=0\)

\(\Leftrightarrow x+7+\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=0\)

=>x+7+1/3-1/10=0

hay x=-217/30

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

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

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

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

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

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

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

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

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

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

Em cảm ơn ạ.

\(2x+1+\frac{1}{6}+1+\frac{1}{12}+..+1+\frac{1}{90}=10\)

=> 2x + 8 + \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=10\)

=> 2x + \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=10-8\)

\(2x+1-\frac{1}{10}=2\)

=> 2x + \(\frac{9}{10}=2\)

=> 2x          = 2 - 9/10

=>2x           = 11/10 

=> x              = 11/10 : 2

x                  =  11/20 

thang Tran ơi,2x+1-1/10 ở đâu vậy

Phải là 2x+1/2-1/10 chứ

mk giải cho nè nhớ tích nha

2x+7/6+13/12+21/20+31/30+43/42+57/56+73/72+91/90=10

2x+(1/6+1)+(1/12+1)+(1/20+1)+(1/30+1)+(1/42+1)+(1/56+1)+(1/72+1)+(1/90+1)=10

2x+(1+1+1+1+1+1+1+1)+(1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)=10

2x+8+(1/2+3+1/3+4+1/4+5+1/5+6+1/6+7+1/7+8+1/8+9+1/9+10)=10

2x+8+(1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)=10

2x+8+(1/2-1/10)=10

2x+8+2/5=10

2x=10-8-2/5

2x=1,6

x=1,6:2

x=0,8

Mk thấy đề bài hơi dư nên x mk tìm ra là số thập phân còn đổi lại ra phân số là 4/5

=9/10-(1/2+1/6+...+1/90)

=9/10-(1-1/2+1/2-1/3+...+1/9-1/10)

=9/10-9/10=0

\(\dfrac{3}{2}+\dfrac{13}{6}+\dfrac{37}{12}+\dfrac{81}{20}+\dfrac{151}{30}+\dfrac{253}{42}+\dfrac{393}{56}+\dfrac{577}{72}+\dfrac{811}{90}\)

\(=\dfrac{459}{10}\)

\(\dfrac{3}{2}+\dfrac{13}{6}+\dfrac{37}{12}+\dfrac{81}{20}+\dfrac{151}{30}+\dfrac{253}{42}+\dfrac{393}{56}+\dfrac{577}{72}+\dfrac{811}{90}\)\(=\dfrac{2}{2}+\dfrac{1}{2}+\dfrac{12}{6}+\dfrac{1}{6}+\dfrac{36}{12}+\dfrac{1}{12}+\dfrac{80}{20}+\dfrac{1}{20}+\dfrac{150}{30}+\dfrac{1}{30}+\dfrac{252}{42}+\dfrac{1}{42}+\dfrac{392}{56}+\dfrac{1}{56}+\dfrac{576}{72}+\dfrac{1}{72}+\dfrac{810}{90}+\dfrac{1}{90}\)\(=1+\dfrac{1}{2}+2+\dfrac{1}{6}+3+\dfrac{1}{12}+4+\dfrac{1}{20}+5+\dfrac{1}{30}+6+\dfrac{1}{42}+7+\dfrac{1}{56}+8+\dfrac{1}{72}+9+\dfrac{1}{90}\)\(=\left(1+2+3+4+5+6+7+8+9\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)= 45 + \(\dfrac{9}{10}\)

= \(\dfrac{459}{10}\)

\(\dfrac{3}{2}+\dfrac{13}{6}+\dfrac{37}{12}+\dfrac{81}{20}+\dfrac{151}{30}+\dfrac{253}{42}+\dfrac{393}{56}+\dfrac{393}{56}+\dfrac{577}{72}+\dfrac{811}{90}\)

\(\dfrac{3780+5460+7770+10206+12684+15180+17685+20195+22708}{2520}\)=\(\dfrac{115668}{2520}=\dfrac{459}{10}\)

2x+7/6+13/12+21/20+31/30+43/42+57/56+73/72+91/90=10

2x+1+1/6+1+1/12+1+1/20+1+1/30+1+1/42+1+1/56+1+1/72+1+1/90=10

2x+(1+1+1+1+1+1+1+1)+(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10)=10

2x+8+(1-1/2+1/2-1/3+...+1/9-1/10)=10

2x+1-1/10=10-8

2x+9/10=2

2x=2-9/10

2x=11/10

x=11/10/2

x=11/20