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4 tháng 12 2015
(x+1/2)+(x+1/4)+(x+1/8)+(x+1/6)=1
(x+x+x+x)+(1/2+1/4+1/6+1/8)=1
4x+(12/24+6/24+4/24+3/24)=1
4x+25/24=1
4x=1-25/24
4x=-1/24
x=-1/24:4
x=-1/96
11 tháng 2 2016
(x+1/2)+(x+1/4)+(x+1/8)+(x+1/16) = (x+x+x+x)+(1/2+1/4+1/8+1/16) = 4x+(8/16+4/16+2/16+1/16) = 4x+15/16 = 1
=> 4x = 1 - 15/16 = 1/16 => x = 1/16 : 4 = 1/64 (= 0,015625)
16 tháng 10 2020
a) \(1+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(=\frac{16}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\)
\(=\frac{23}{16}\)
b) \(2-\frac{1}{8}-\frac{1}{12}-\frac{1}{16}\)
\(=\frac{96}{48}-\frac{6}{48}-\frac{4}{48}-\frac{3}{48}\)
\(=\frac{83}{48}\)
c) \(\frac{4}{99}\cdot\frac{18}{5}\div\frac{12}{11}+\frac{3}{5}\)
\(=\frac{4\cdot18\cdot11}{99\cdot5\cdot12}+\frac{3}{5}\)
\(=\frac{4\cdot9\cdot2\cdot11}{9\cdot11\cdot5\cdot4\cdot3}+\frac{3\cdot3}{3\cdot5}\)
\(=\frac{2}{15}+\frac{9}{15}=\frac{11}{15}\)
d) \(\left(1-\frac{3}{4}\right)\left(1+\frac{1}{3}\right)\div\left(1-\frac{1}{3}\right)\)
\(=\frac{1}{4}\cdot\frac{4}{3}\div\frac{2}{3}\)
\(=\frac{1\cdot4\cdot3}{4\cdot3\cdot2}=\frac{1}{2}\)
11 tháng 9 2019
Bài 1 : \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right]:5\times x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{1}{24}+\frac{2}{15}+\frac{3}{40}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{5}{12}:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{1}{12}\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{x}{12}< \frac{5}{6}\)
=> \(\frac{8}{12}< \frac{x}{12}< \frac{10}{12}\)
=> x = 9
Bài 2 : \(\frac{\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right]}{x}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
=> \(\frac{\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right]}{x}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)
=> \(\frac{\left[1-\frac{1}{16}\right]}{x}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=1-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=\frac{11}{12}\)
=> \(\frac{15}{16}:x=\frac{11}{12}\)
=> \(x=\frac{45}{44}\)
Bài 3 : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times(x+1):2}=\frac{399}{400}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times(x+1)}=\frac{399}{400}\)
=> \(2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(\left[\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{399}{800}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
=> \(\frac{1}{x+1}=\frac{1}{800}\)
=> x = 799
11 tháng 9 2019
Bài 2 :
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\) (*)
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}=\frac{8+4+2+1}{16}=\frac{15}{16}\) (1)
Lại có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=1\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)
\(=1-\frac{1}{12}=\frac{11}{12}\) (2)
Thay (1) và (2) vào biểu thức (*) ta được :
\(\frac{15}{16}:x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{16}:\frac{11}{12}\)
\(\Leftrightarrow x=\frac{45}{44}\)
Vậy : \(x=\frac{45}{44}\)