Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{6}+\frac{1}{32}\)
\(\frac{4+2+1}{8}+\frac{1}{6}+\frac{1}{32}\)
\(\frac{9}{8}+\frac{1}{6}+\frac{1}{32}\)
\(\frac{9}{8}+\frac{1}{32}+\frac{1}{6}\)
\(\frac{36+32}{32}+\frac{1}{6}\)
\(\frac{68}{32}+\frac{1}{6}=\frac{17}{8}+\frac{1}{6}\)
\(\frac{51+4}{24}=\frac{55}{24}\)
a, \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
= \(1-\frac{1}{7}=\frac{6}{7}\)
\(M=1-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)\)
\(=1-\left(\dfrac{3}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)\)
\(=1-\left(\dfrac{7}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)\)
\(=1-\left(\dfrac{15}{16}+\dfrac{1}{32}\right)\)
\(=1-\dfrac{31}{32}=\dfrac{1}{32}\)
b: A=1/3+1/9+...+1/3^10
=>3A=1+1/3+...+1/3^9
=>A*2=1-1/3^10=(3^10-1)/3^10
=>A=(3^10-1)/(2*3^10)
c: C=3/2+3/8+3/32+3/128+3/512
=>4C=6+3/2+...+3/128
=>3C=6-3/512
=>C=1023/512
d: A=1/2+...+1/256
=>2A=1+1/2+...+1/128
=>A=1-1/256=255/256
a)
\(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(\frac{2.3\left(1.2\right)+2.3\left(2.4\right)+2.3\left(3.6\right)+2.3\left(4.8\right)+2.3\left(5.10\right)}{3.4\left(3.4+6.8+9.12+12.16+15.20\right)}\)
\(=\frac{\left(3.4+6.8+9.12+12.16+15.20\right)}{2.3\left(3.4+6.8+9.12+12.16+15.20\right)}=\frac{1}{2.3}=\frac{1}{6}\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{1}{32}=\dfrac{48+24+12+16+3}{96}=\dfrac{103}{96}.\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{6}+\dfrac{1}{32}=\dfrac{1.48}{2.48}+\dfrac{1.24}{4.24}+\dfrac{1.12}{8.12}+\dfrac{1.16}{6.16}+\dfrac{1.3}{32.3}=\dfrac{48+24+12+16+3}{96}=\dfrac{103}{96}\)