K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
CP
2
NK
0
EC
2
NN
1
16 tháng 7 2016
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
NQ
0
12 tháng 9 2017
\(x+\frac{13}{12}+\frac{21}{20}+...+\frac{91}{90}\)
\(=x+1+\frac{1}{12}+1+\frac{1}{20}+...+1+\frac{1}{90}\)
\(=x+\left(1+1+1+1+1+1+1\right)+\left(\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=x+7+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=x+7+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=x+7+\left(\frac{1}{3}-\frac{1}{10}\right)\)
\(=x+7+\frac{7}{30}\)
\(=x+7\frac{7}{30}\)
Q
15 tháng 4 2017
\(2x+\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}=10\)
\(2x+\left(1+\dfrac{1}{6}\right)+\left(1+\dfrac{1}{12}\right)+\left(1+\dfrac{1}{20}\right)+\left(1+\dfrac{1}{30}\right)+\left(1+\dfrac{1}{42}\right)+\left(1+\dfrac{1}{56}\right)+\left(1+\dfrac{1}{72}\right)+\left(1+\dfrac{1}{90}\right)=10\)
\(2x+8+\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=10\)
\(2x+8+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)=10\)
\(2x+8+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)=10\)
\(2x+8+\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=10\)
~ Chúc bạn học tốt ~
15 tháng 4 2017
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
11 tháng 7 2018
\(2x+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}+\frac{91}{90}=10\)
=> \(2x+\frac{6+1}{6}+\frac{12+1}{12}+....+\frac{90+1}{90}=10\)
=> \(2x+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+10=10\)
=> \(2x+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=0\)
=>\(2x+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}=0\)
=>\(2x+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=0\)
=> \(2x-\frac{1}{10}=0\)
=>2x=\(\frac{1}{10}\)=> x=1/20
11 tháng 7 2018
mình có bị nhầm chỗ dấu suy ra thứ 3. đáng lẽ ra biểu thức đó cộng 8 chứ k phải cộng 10 do mình sơ ý nên bạn hãy sủa lại chỗ ấy
\(M=\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}\)
\(\Rightarrow M=\left(1+\frac{1}{12}\right)+\left(1+\frac{1}{20}\right)+\left(1+\frac{1}{30}\right)+\left(1+\frac{1}{42}\right)+\left(1+\frac{1}{56}\right)+\left(1+\frac{1}{72}\right)\)
\(\Rightarrow M=6+\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)\)
\(\Rightarrow M=6+\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\)
\(\Rightarrow M=6+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(\Rightarrow M=6+\left(\frac{1}{3}-\frac{1}{9}\right)=6+\left(\frac{3}{9}-\frac{1}{9}\right)\)
\(\Rightarrow M=6+\frac{2}{9}=\frac{54}{9}+\frac{2}{9}=\frac{56}{9}\)