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Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
1/5.7 + 1/7.9 + 1/9.11 + ... + 1/49.51
= 1/2 . (2/5.7 + 2/7.9 + 2/9.11 + ... + 2/49.51)
= 1/2 . (1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/49 - 1/51)
= 1/2 . (1/5 - 1/51)
= 1/2 . 46/255
= 23/255
S = \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...\frac{1}{49}-\frac{1}{51}\)
S = \(\frac{1}{5}-\frac{1}{51}=\frac{46}{255}\)
a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
=\(\frac{1}{2}x\left(\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}+...+\frac{2}{2015x2017}\right)\)
=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{2017}\right)\)
=\(\frac{1}{2}x\frac{2012}{10085}\)
=\(\frac{1006}{10085}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
Đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{1}{1}-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}\div2\)
\(\Rightarrow A=\frac{50}{101}\)
ta có B= 1/31+1/32+1/33+...+1/60
=> B=(1/30+1/30+...+1/30) + (1/40+1/40+1/40+...+1/40)
10 số hạng 10 số hạng
=> B< 10/30+10/40+10/50
=> = 1/3+1/4+1/5
=> = 47/60
=> B< 47/60 < 48/60= 4/5
Vế 2 tự làm nha bà
A)
\(\frac{1}{30}\)-\(\frac{1}{31}\)+\(\frac{1}{31}\)-\(\frac{1}{32}\)+\(\frac{1}{32}\)-\(\frac{1}{33}\)+...+\(\frac{1}{42}\)-\(\frac{1}{43}\)
=\(\frac{1}{30}\)-\(\frac{1}{43}\)
=\(\frac{13}{1290}\)
B)
=\(\frac{2}{2}\)X(\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+\(\frac{1}{7.9}\)+\(\frac{1}{9.11}\))
=\(\frac{1}{2}\)X(\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+\(\frac{2}{9.11}\))
=\(\frac{1}{2}\)X(\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))
=\(\frac{1}{2}\)X(\(\frac{1}{3}\)-\(\frac{1}{11}\))
=\(\frac{1}{2}\)X\(\frac{8}{33}\)
=\(\frac{8}{66}\)=\(\frac{4}{33}\)