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

=> 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

\(\frac{-21}{5}\)

x= -127/30 .

gợi ý các bạn chọn ( k) đúng cho mình 

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

\(x+\frac{13}{12}+\frac{21}{20}+...+\frac{91}{90}\)

\(=x+1+\frac{1}{12}+1+\frac{1}{20}+...+1+\frac{1}{90}\)

\(=x+\left(1+1+1+1+1+1+1\right)+\left(\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=x+7+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)

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

\(=x+7+\left(\frac{1}{3}-\frac{1}{10}\right)\)

\(=x+7+\frac{7}{30}\)

\(=x+7\frac{7}{30}\)

7va7/30

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

a) 

\(=\frac{7\cdot7\cdot8\cdot8\cdot9\cdot9\cdot10\cdot10\cdot11\cdot11}{6\cdot8\cdot7\cdot9\cdot8\cdot10\cdot9\cdot11\cdot10\cdot12}\)    

\(=\frac{7\cdot11}{6\cdot12}\)     

\(=\frac{77}{72}\)

b) 

\(=1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)  

\(=6+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)  

\(=6+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)  

\(=6+\frac{1}{2}-\frac{1}{8}\)  

\(=6+\frac{3}{8}\)

\(=\frac{51}{8}\)

Chia thành...a và b nhé.

Bg

a)Ta có: \(\frac{49}{48}.\frac{64}{63}.\frac{81}{80}.\frac{100}{99}.\frac{121}{120}\)

= \(\frac{49.64.81.100.121}{48.63.80.99.120}\)

= \(\frac{7.7.8.8.9.9.10.10.11.11}{6.8.7.9.8.10.9.11.10.12}\)

= \(\frac{7.11}{6.12}\)    (chịt tiêu trên dưới)

= \(\frac{77}{72}\)

b) Ta có: \(\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}\)

Có 6 số hạng  (đếm)

= \(1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)

= \(1+1+...+1+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

= \(1.6+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

= \(6+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

= \(6+\frac{1}{2}-\frac{1}{8}\)

= \(\frac{13}{2}-\frac{1}{8}\)

= \(\frac{51}{8}\)

Hơi dài....

THANK YOU