1/6 + 1/12 + 1/24 + ... + 1/192 giúp mình với!
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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
= \(\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{12}+\frac{1}{24}\right)+\left(\frac{1}{48}+\frac{1}{96}\right)+\frac{1}{192}\)
= \(\left(\frac{1}{2}+\frac{1}{8}\right)+\left(\frac{1}{32}+\frac{1}{192}\right)\)
= \(\frac{5}{8}+\frac{1}{192}\)
= \(\frac{121}{192}\)
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
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
1/3 + 1/6 + 1/12 + 1/24 + 1/48 + 1/96 + 1/192
= 127/192
k minh di , xin day
\(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+...+\dfrac{1}{192}\)
\(2A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{96}\)
\(A=\dfrac{1}{3}-\dfrac{1}{192}=\dfrac{64}{192}-\dfrac{1}{192}=\dfrac{63}{192}=\dfrac{21}{64}\)