\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

= \(\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{12}+\frac{1}{24}\right)+\left(\frac{1}{48}+\frac{1}{96}\right)+\frac{1}{192}\)

=               \(\left(\frac{1}{2}+\frac{1}{8}\right)+\left(\frac{1}{32}+\frac{1}{192}\right)\)

=                              \(\frac{5}{8}+\frac{1}{192}\)

=                                    \(\frac{121}{192}\)

ko ai bết đâu

=\(\dfrac{64}{192}+\dfrac{32}{192}+\dfrac{16}{192}+\dfrac{8}{192}+\dfrac{4}{192}+\dfrac{2}{192}+\dfrac{1}{192}\)

= \(\dfrac{127}{192}\)

Giải:

a) \(\dfrac{-5}{8}=\dfrac{x}{16}\) 

\(\Rightarrow x=\dfrac{16.-5}{8}=-10\) 

\(\dfrac{3x}{9}=\dfrac{2}{6}\) 

\(\Rightarrow3x=\dfrac{2.9}{6}=3\) 

\(\Rightarrow x=1\)

b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)  

\(\Rightarrow x+3=\dfrac{1.15}{3}=5\) 

\(\Rightarrow x=2\)

\(\dfrac{6}{2x+1}=\dfrac{2}{7}\) 

\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\) 

\(\Rightarrow x=10\)

c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\) 

\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\) 

\(\Rightarrow x=0\) 

\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow y=\dfrac{-12.24}{18}=-16\) 

 \(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\) 

\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\) 

\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\) 

\(\Rightarrow x=-29\) 

\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\) 

\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\) 

d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\) 

\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\) 

\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\) 

\(\Rightarrow x\in\left\{-3;-2;-1\right\}\) 

\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\) 

\(\Rightarrow x\in\left\{-1;0;1;2\right\}\) 

e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\) 

\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\) 

\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\) 

\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\) 

\(\Rightarrow5x+230=100x+40\) 

\(\Rightarrow5x-100x=40-230\) 

\(\Rightarrow-95x=-190\) 

\(\Rightarrow x=-190:-95\) 

\(\Rightarrow x=2\) 

\(y\dfrac{5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow y^2+5=86\) 

\(\Rightarrow y^2=86-5\) 

\(\Rightarrow y^2=81\) 

\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\) 

Chúc bạn học tốt!

a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)

 =  \(\dfrac{5}{6}\)  + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)

= 1 + \(\dfrac{1}{23}\)

 = \(\dfrac{24}{23}\) 

b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)

= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5

= 1 - 1 + 0,5

= 0,5 

c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)

=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)

=0

d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)

= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)

= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)

= 1

e, 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)

 =    (0,5 + 0,4) + ( \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\)) + (\(\dfrac{5}{7}\)  - \(\dfrac{4}{35}\))

 = 0,9 + \(\dfrac{1}{2}\) + \(\dfrac{3}{5}\)

= 0,9 + 0,5 + 0,6

= 2

1/3 + 1/6 + 1/12 + 1/24 + 1/48 + 1/96 + 1/192 + 1/384 = 85/128

A=1/2+1/6+1/12+1/24+…+1/192+1/384

Giải

Ta có:

A x 2 =(1/2+1/6+1/12+1/24+…+1/192+1/384) x 2

A x 2 = 1 + 2/6 + 2/12 + 2/24 + ... + 2/192 + 2/384

Rút gọn ta được:

A x 2 = 1 + 1/3 + 1/6 + 1/12 + ... + 1/96 + 1/192

A x 2 - A = 1 + 1/3 + 1/6 + 1/12 + ... + 1/96 + 1/192 - (1/2+1/6+1/12+1/24+…+1/192+1/384)

A = 1 + 1/3 - 1/2 - 1/384

A = 5/6 - 1/384

A = 319/384

ĐS: 319/384

Sao đề bài là 1/3 còn khi làm là1/2 hả

1/3 + 1/6 + 1/12 + 1/24 + 1/48 + 1/96 + 1/192

= 127/192

k minh di , xin day