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\(A=\left(1+\frac{1}{2}\right)x\left(1+\frac{1}{3}\right)x\left(1+\frac{1}{4}\right)x...x\left(1+\frac{1}{100}\right)\)
\(A=\frac{3}{2}x\frac{4}{3}x\frac{5}{4}x...x\frac{101}{100}\)
\(A=\frac{101}{2}\)
A = \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{101}{100}\)
A = \(\frac{101}{2}\)
Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{8}\right)\)
\(\Rightarrow A=2\cdot\frac{3}{8}=\frac{3}{4}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{96}\)
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{48}\)
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{24}\)
...
\(A+\frac{1}{96}=\frac{1}{3}+\frac{1}{3}\Rightarrow A=\frac{2}{3}-\frac{1}{96}=\frac{2\cdot32-1}{96}=\frac{63}{96}=\frac{21}{32}\).
MSC:192
\(\frac{4}{3}+\frac{1}{3}+\frac{1}{12}+\frac{1}{48}+\frac{1}{192}\)
\(=\frac{256}{192}+\frac{64}{192}+\frac{16}{192}+\frac{4}{192}+\frac{1}{192}\)
\(=\frac{341}{192}\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)
=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)
tính nhanh hộ mình câu này với :
\(\frac{4}{3}+\frac{1}{3}+\frac{1}{12}+\frac{1}{48}+\frac{1}{192}\)
= 1 - 4/3 + 1/3 - 1/3 + 1/12 - 1/12 + 1/48 - 1/48 + 1/92
= 1 + 1/92
= 92/92 + 1/92
= 93/92
Ko biết có đúng không nữa!
\(\frac{4}{3}+\frac{1}{3}+\frac{1}{3x4}+\frac{1}{3x4^2}+\frac{1}{3x4^3}=\frac{4^4+1x4^3+1x4^2+1x4+1}{3x4^3}.\)
\(=\frac{256+64+16+4+1}{3x4^3}=\frac{341}{192}\)
Đặt: \(A=\frac{1}{14}+\frac{1}{28}+\frac{1}{56}+....+\frac{1}{896}\)
\(\Rightarrow\)\(2A=\frac{1}{7}+\frac{1}{14}+\frac{1}{28}+....+\frac{1}{448}\)
\(\Rightarrow\)\(2A-A=\left(\frac{1}{7}+\frac{1}{14}+....+\frac{1}{448}\right)-\left(\frac{1}{14}+\frac{1}{28}+.....+\frac{1}{896}\right)\)
\(\Rightarrow\)\(A=\frac{1}{7}-\frac{1}{896}=\frac{127}{896}\)
mà \(A=x-3\)
nên \(x-3=\frac{127}{896}\)
\(\Rightarrow\)\(x=3\frac{127}{896}\)