\(x\) :

\(\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...">

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16 tháng 7 2016

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{9}\right)\)

\(=\frac{1}{2}.\frac{8}{9}\)

\(=\frac{4}{9}\)

16 tháng 7 2016

Đặt: A=1/1.3+1/3.5+1/5.7+1/7.9

2A=2/1.3+2/3.5+2/5.7+2/7.9

2A=1-1/3+1/3-1/5+1/5-1/7+1/7-1/9

2A=1-1/9

2A=8/9

A=4/9

13 tháng 7 2018

a) (1,5 . 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125

=> (2,85 - x - 0,5) : 0,25 = 60

=> (2,85 - 0,5) - x = 60 . 0,25

=> 2,35 - x = 15

=> x = 2,35 - 15

=> x = -12,65

Vậy x = -12,65

b) \(1-\left(5\frac{2}{9}+x-7\frac{7}{18}\right)\div2\frac{1}{6}=0\)

\(\Rightarrow\left(5\frac{2}{9}-7\frac{7}{18}+x\right)\div2\frac{1}{6}=1-0\)

\(\Rightarrow\left(\frac{47}{9}-\frac{133}{18}+x\right)\div2\frac{1}{6}=1\)

\(\Rightarrow\frac{-13}{6}+x=2\frac{1}{6}\)

\(\Rightarrow x=2\frac{1}{6}-\frac{-13}{6}\)

\(\Rightarrow x=\frac{13}{6}+\frac{13}{6}\)

\(\Rightarrow x=\frac{26}{6}\)

\(\Rightarrow x=\frac{13}{3}\)

Vậy \(x=\frac{13}{3}\)

c) \(35\left(2\frac{1}{5}-x\right)=32\)

\(\Rightarrow2\frac{1}{5}-x=32\div35\)

\(\Rightarrow\frac{11}{5}-x=\frac{32}{35}\)

\(\Rightarrow x=\frac{11}{5}-\frac{32}{35}\)

\(\Rightarrow x=\frac{9}{7}\)

Vậy \(x=\frac{9}{7}\)

d) \(\frac{4}{3}+\left(x\div2\frac{2}{3}-0,5\right).1\frac{35}{55}=0,6\)

\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{3}{5}-\frac{4}{3}\)

\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{-11}{15}\)

\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-11}{15}\div\frac{18}{11}\)

\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-121}{270}\)

\(\Rightarrow x\div\frac{8}{3}=\frac{-121}{270}+\frac{1}{2}\)

\(\Rightarrow x\div\frac{8}{3}=\frac{7}{135}\)

\(\Rightarrow x=\frac{7}{135}.\frac{8}{3}\)

\(\Rightarrow x=\frac{56}{405}\)

Vậy \(x=\frac{56}{405}\)

e) \(1\frac{1}{3}.2\frac{2}{4}\div\frac{5}{6}.1\frac{1}{11}=11-5\div x\)

\(\Rightarrow\frac{4}{3}.\frac{5}{2}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)

\(\Rightarrow\frac{10}{3}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)

\(\Rightarrow4.\frac{12}{11}=11-5\div x\)

\(\Rightarrow11-5\div x=\frac{48}{11}\)

\(\Rightarrow5\div x=11-\frac{48}{11}\)

\(\Rightarrow5\div x=\frac{73}{11}\)

\(\Rightarrow x=5\div\frac{73}{11}\)

\(\Rightarrow x=\frac{55}{73}\)

Vậy \(x=\frac{55}{73}\)

a) (1,5 * 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125

(2,85 - x - 0,5) : 0,25 = 60

(2,85 - x - 0,5) = 60 x 0,25

(2,85 - x - 0,5) = 15

2,35 - x = 15

x = 2,35 - 15

x = -12,65

\(\left(a\right)\frac{34-x}{30}=\frac{5}{6}\)

\(\frac{34-x}{30}=\frac{25}{30}\)

34 - x = 25

x = 34 - 25 = 9

\(\left(b\right)\frac{x+13}{34}=\frac{12}{17}\)

\(\frac{x+13}{34}=\frac{24}{34}\)

x + 13 = 24

x = 24 - 13 = 11

\(\left(c\right)\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{56}{81}\)

\(4x+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)

Đặt \(A=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)

Ta có : \(3A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)

\(3A-A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}-\frac{1}{3}-\frac{1}{9}-\frac{1}{27}-\frac{1}{81}\)

\(2A=1-\frac{1}{81}=\frac{80}{81}\)

\(A=\frac{80}{81}\div2=\frac{40}{81}\)

\(\Rightarrow4x+\frac{40}{81}=\frac{56}{81}\)

\(4x=\frac{56}{81}-\frac{40}{81}\)

\(4x=\frac{16}{81}\)

\(x=\frac{16}{81}\div4=\frac{4}{81}\)

9 tháng 8 2020

a, \(\frac{34-x}{30}=\frac{5}{6}\Leftrightarrow\frac{34-x}{30}=\frac{25}{30}\)

\(\Leftrightarrow34-x=25\Leftrightarrow x=9\)

b, \(\frac{x+13}{34}=\frac{12}{17}\Leftrightarrow\frac{x+13}{34}=\frac{24}{34}\)

\(\Leftrightarrow x+13=24\Leftrightarrow x=11\)

28 tháng 8 2017

Ta co  \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+....+\frac{2}{x\cdot\left(x+1\right)}\)

\(\Rightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\cdot\left(x+1\right)}=\frac{2}{9}\)

\(\Rightarrow\)\(2\cdot\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Rightarrow\)\(2\cdot\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Rightarrow\)\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)

\(\Leftrightarrow x+1=18\)

       \(x=17\)

12 tháng 8 2018

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