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\(\frac{4}{1}.\frac{6}{2}.\frac{8}{3}.\frac{10}{4}...\frac{64}{31}=\frac{2^{31}.\left(2.3.4.5...32\right)}{1.2.3.4...31}=2^{31}.32\)
Mà \(2^x=2^{31}.32=2^{31}.2^5=2^{36}\)
\(\Rightarrow x=36\)
\(E=\frac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}=\frac{1\cdot1\cdot1\cdot1\cdot.....\cdot1\cdot1}{2\cdot2\cdot2\cdot....\cdot2\cdot64}=\frac{1}{2\cdot30\cdot64}=\frac{1}{3840}\)
c)\(\frac{1}{2}x+\frac{1}{8}x=\frac{3}{4}\)
\(\Rightarrow x.\left(\frac{1}{2}-\frac{1}{8}\right)=\frac{3}{4}\)
\(\Rightarrow x.\frac{3}{8}=\frac{3}{4}\)
=>x\(=\frac{3}{4}:\frac{3}{8}\)
=>x=\(2\)
a)\(x+\frac{1}{6}=\frac{-3}{8}\)
=>\(x=\frac{-3}{8}-\frac{1}{6}\)
=>\(x=\frac{-9}{24}-\frac{4}{24}\)
=>\(x=\frac{-13}{24}\)
b)\(2-\left|\frac{3}{4}-x\right|=\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=2-\frac{7}{12}\)
=>\(\left|\frac{3}{4}-x\right|=\frac{24}{12}-\frac{7}{12}\)
\(\Rightarrow\left|\frac{3}{4}-x\right|=\frac{17}{12}\)
TH1: \(\frac{3}{4}-x=\frac{17}{12}\)
=>x=\(\frac{3}{4}-\frac{17}{12}\)
=>x=\(x=-\frac{2}{3}\)
TH2:\(\frac{3}{4}-x=-\frac{17}{12}\)
=>\(x=\frac{3}{4}-\left(-\frac{17}{12}\right)\)
=>x=\(x=\frac{13}{6}\)
Dzồi nhìu phết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\) <=>\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}+.....\frac{1}{n}-\frac{1}{n-1}\right)=\frac{1999}{2001}\)
<=>\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2001}\)
<=>\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
<=>\(\frac{1}{n+1}=\frac{1}{2001}\)
<=>n+1 =2001
<=>n = 2000
ta có:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}\right)=\frac{1}{2}.\frac{1999}{2001}\)
\(\frac{1}{2.3}+\frac{1}{2.6}+\frac{1}{2.10}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n\left(n+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2001}\)
=>\(n+1=2001\)
=>\(n=2000\)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
Ta có\(\frac{-2}{3}\)+\(\frac{1}{4}\)= \(\frac{-8}{12}\)+\(\frac{3}{12}\)= \(\frac{-5}{12}\)
\(\frac{3}{4}\)-\(\frac{1}{3}\)=\(\frac{9}{12}\)-\(\frac{4}{12}\)=\(\frac{5}{12}\)
=> \(\frac{-5}{12}\)<\(\frac{a}{6}\)<\(\frac{5}{12}\)
=> \(\frac{-5}{12}\)<\(\frac{2a}{12}\)<\(\frac{5}{12}\)
Mà a là số nguyên,2a là số chẵn
=>2a
=>a
Có thể đề là: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.....\frac{31}{64}=2^n\)
=> \(\frac{1.2.3.4....31}{\left(2.2\right)\left(2.3\right).\left(2.3\right)\left(2.4\right)\left(2.5\right)...\left(2.31\right).\left(2.32\right)}=2^n\)
=> \(\frac{1.2.3.4...31}{2^{16}.\left(2.3.4.5..31.32\right)}=2^n\) => \(\frac{1}{2^{16}.32}=2^n\) => 216.25.2n = 1
=> 231+n = 1 = 20 => 31 + n = 0 => n = -31
\(=\frac{1.2.3.....30}{4.6.8....64}=\frac{1}{2.2...2.64}=\frac{1}{2^{30}.2^6}=\frac{1}{2^{36}}\) ( 30 số 2)
=> 2^n = 1/2^36
=> n = -36