Tính nhanh:
a, \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{201\times203}\)
b, \(\dfrac{-4}{2.5}-\dfrac{4}{5.8}-\dfrac{4}{8.11}-...-\dfrac{4}{2015.2018}\)
So sánh:
A = \(\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+\dfrac{3^2}{8.11}\) và B = \(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
So sánh: A =\(\dfrac{3^2}{2.5}+\dfrac{3^2}{5.8}+\dfrac{3^2}{8.11}\) và B \(\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{59.61}\)
tính biểu thức : A = \(\dfrac{1}{2.5}\)+ \(\dfrac{1}{5.8}\)+ \(\dfrac{1}{8.11}\)+...... + \(\dfrac{1}{2012.2015}\)+ \(\dfrac{1}{2015.2018}\)+\(\dfrac{1}{2018.2021}\)
3A=3/2.5+...+3/2018.2021
3A=1/2-1/5+1/5-...+1/2018-1/2021
3A=1/2-1/2021 sau tự tính A
3A= 1/2- 1/5 + 1/5- 1/8+ 1/8 -1/11+...+ 1/2012- 1/2015 +1/2015- 1/2018-1/2021
3A =1/2 -1/2021
3A = 2019/ 4042
=> 2019/4042 : 3 = 673/4042
Chúc bạn học tốt !!
Giải:
A=1/2.5+1/5.8+1/8.11+...+1/2012.2015+1/2015.2018+1/2018.2021
A=1/3.(3/2.5+3/5.8+3/8.11+...+3/2012.2015+3/2015.2018+3/2018.2021)
A=1/3.(1/2-1/5+1/5+1/8+1/8-1/11+...+1/2012-1/2015+1/2015-1/2018+1/2018-1/2021)
A=1/3.(1/2-1/2021)
A=1/3.2019/4042
A=673/4042
Chúc bạn học tốt!
Tính nhanh:
M= \(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
B = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{99.101}\)
\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(B=2.\frac{100}{101}=\frac{200}{101}\)
\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)
\(X=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{99^2}{98.100}\)
\(K=\dfrac{1}{3}.\dfrac{1}{15}.\dfrac{1}{35}...\dfrac{1}{9999}\)
\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\dfrac{96}{505}\)
\(\Rightarrow G=\dfrac{64}{505}\)
\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\\ G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{101}{505}-\dfrac{5}{505}\right)\\ G=\dfrac{2}{3}.\dfrac{96}{505}\\ G=\dfrac{64}{505}\)
Tính nhanh:
M= \(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
B = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{99.101}\)
Cứu mai thi rồi
Ta có :
M= \(\dfrac{3+3-3+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4+4-4+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)= \(\dfrac{3+3-3}{4+4-4}=\dfrac{3}{4}\)
b) Nhận xét thấy: \(\dfrac{2}{1.3}=1-\dfrac{1}{3};\dfrac{1}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};...\)
Ta có:
B= 1-\(\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
B= 1- \(\dfrac{1}{101}\)= \(\dfrac{100}{101}\)
Vậy B= \(\dfrac{100}{101}\)
Bài 7: A=\(\dfrac{4}{2.5}\)+\(\dfrac{4}{5.8}\)+\(\dfrac{4}{8.11}\)+...+\(\dfrac{4}{65.68}\)
Ta có: \(A=\dfrac{4}{2\cdot5}+\dfrac{4}{5\cdot8}+...+\dfrac{4}{65\cdot68}\)
\(=\dfrac{4}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{65\cdot68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\left(\dfrac{1}{2}-\dfrac{1}{68}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{68}=\dfrac{11}{17}\)
1/.\(\dfrac{1}{1}.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{5}\)
2/.\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{10100}\)
3/.A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
4/.A = \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
tính bằng cách thuận tiện nhất ( làm nhanh trước 5h nha , nếu ai làm được thì cho 100 tick , thật đó và trình bày cách diễn giải nha )
2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)
= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)
=1-\(\dfrac{1}{101}\)=...........
mk làm vậy thôi nha
thông cảm
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)
=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)
tương tự
Tính giá trị biểu thức:
B= \(\dfrac{\left(-2\right)^{24}.3^5-4^{12}.9^2}{8^8.3^5}+\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{301.303}\)
\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)
\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)
\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)
=168/101