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\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{182}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Mình chỉnh lại đề B nha:
\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
: A = 1/6+1/12+1/20+1/30+.........+1/210
A = 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/14.15
A = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + ... + 1/14 - 1/15
A = 1/2 - 1/15
A = 13/30
Ta có: 1/2= 1/1- 1/2
1/6= 1/2 - 1/3
1/12= 1/3- 1/4
...
1/30= 1/5 - 1/6
1/42= 1/6 - 1/7
Thay vào tổng kia: 1/2+1/6+...+1/30+1/42= 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/5 - 1/6 + 1/6 - 1/7 = 1/2 - 1/7= 5/14
Chúc bạn học tốt. Thân!
\(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{420}\\ \Rightarrow A=\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{20\times21}\\ \Rightarrow A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\\\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{20}=\dfrac{9}{20}\)
Sau khi rút gọn phải còn:
\(A=\dfrac{1}{2}-\dfrac{1}{21}\) (chứ anh)
\(A=\dfrac{19}{42}\)
Có công thức \(\dfrac{x}{a\left(a+x\right)}=\dfrac{1}{a}-\dfrac{1}{a+x}\) nhé!
Ví dụ: \(\dfrac{2}{2.4}=\dfrac{1}{2}-\dfrac{1}{4}\)
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(=1-\dfrac{1}{8}=\dfrac{7}{8}\)
Dấu . tức là nhân nhé!
\(\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)
= 1 + 2 + 2
= 5
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
=\(1-\frac{1}{7}\)
=\(\frac{6}{7}\)
a) (1/15 + 1/35 + 1/65) . x = 1
=> 151/1365 . x = 1
=> x = 1 : 151/1365
=> x = 1365/151
b) (1/2 + 1/4 + 1/8 + 1/16) : x = 1/2 + 1/6 +...+1/132
=> 15/16 : x = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)
=> 15/16 : x = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)
=> 15/16 : x = \(1-\frac{1}{12}\)
=> 15/16 : x = \(\frac{11}{12}\)
=> x = \(\frac{15}{16}:\frac{11}{12}\)
=> x = \(\frac{45}{44}\)
Ta có:
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(\Rightarrow2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)=1+\frac{1}{2}+...+\frac{1}{2^8}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)\)
\(\Rightarrow A=1-\frac{1}{2^9}=1-\frac{1}{512}=\frac{511}{512}\)
Vậy giá trị biểu thức là \(\frac{511}{512}\)
b) Ta có:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}+\frac{1}{110}=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
Vậy giá trị biểu thức là \(\frac{10}{11}\)