1+1+1+1+1+×1×1×1=?????giúp mình nha
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\(F=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right)...\left(1-\frac{1}{100}\right)\)
\(F=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{99}{100}\)
\(F=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{9.11}{10.10}\)
\(F=\frac{1.2.3.4...9}{2.3.4...10}.\frac{3.4.5...11}{2.3.4.5...10}\)
\(F=\frac{1}{10}.\frac{11}{2}=\frac{11}{21}\)
\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)
\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)
\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(A=\frac{1}{2}\times\frac{98}{303}\)
\(A=\frac{49}{303}\)
A= \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ ... + \(\frac{1}{9999}\)
A= \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\) + ... + \(\frac{1}{99.101}\)
2. A= \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) + ... + \(\frac{2}{99.101}\)
2.A = \(\frac{1}{3}\) - \(\frac{1}{5}\)+ \(\frac{1}{5}\)-\(\frac{1}{7}\) + ... + \(\frac{1}{99}\) - \(\frac{1}{101}\)
2.A= \(\frac{1}{3}\) - \(\frac{1}{101}\)
2.A= \(\frac{101}{303}\) - \(\frac{3}{303}\)
2.A= \(\frac{98}{303}\)
A = \(\frac{98}{303}\) : 2
A = \(\frac{49}{303}\)
Vay A=\(\frac{49}{303}\)
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{10^2}\right)\)
= \(\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}.\dfrac{4^2-1}{4^2}...\dfrac{10^2-1}{10^2}\)
= \(\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{9.11}{10^2}\)
= \(\dfrac{\left(1.2.3...9\right).\left(3.4.5...11\right)}{\left(2.3.4...10\right)\left(2.3.4...10\right)}\)
= \(\dfrac{1.11}{10.10}=\dfrac{11}{100}\)
1+1-1+1-1+1-1+1-1+6+199999999999999999999999999999999999 = 200000000000000000000000000000000006
1+1-1+1-1+1-1+1-1+6+199999999999999999999999999999999999=200000000000000000000000000000000006
A = \(-\dfrac{1}{20}\) + \(\dfrac{-1}{30}\) + \(\dfrac{-1}{42}\) + \(\dfrac{-1}{56}\) + \(\dfrac{-1}{72}\) + \(\dfrac{-1}{90}\)
A = - ( \(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\))
A = - ( \(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\))
A = - ( \(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\))
A = - (\(\dfrac{1}{4}-\dfrac{1}{10}\))
A = - \(\dfrac{3}{20}\)
F=3/4*8/9*15/16*24/25*...*9*10
F=\(\frac{3\cdot8\cdot15\cdot24\cdot...9}{4\cdot9\cdot16\cdot25\cdot...10}\)
F=\(\frac{3\cdot2\cdot15\cdot6}{6\cdot3\cdot8\cdot25}\)
F-\(\frac{2\cdot15}{8\cdot25}\)
F=\(\frac{3}{20}\)
=5 nhé bạn
Ai giúp mk mk k cho