Cho \(B=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+...............+\dfrac{99^2}{98.100}\) Tìm phần nguyên của B
Help me!!!
\(\dfrac{2^2}{1.3}\)x\(\dfrac{3^2}{2.4}\)x\(\dfrac{4^2}{3.5}\)......\(\dfrac{99^2}{98.100}\)
\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{99^2}{98.100}\)
\(=\dfrac{2.2.3.3.4.4.....99.99}{1.3.2.4.3.5.....98.100}\)
\(=\dfrac{2.3.4.....99}{1.2.3.4.....98}.\dfrac{2.3.4.....99}{3.4.5.....100}\)
\(=\dfrac{99}{98}\cdot\dfrac{2}{100}\)
\(=\dfrac{99}{4900}\)
Tính: \(B=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{98.100}{99^2}\)
Ta có:B = \(\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}......\dfrac{98.100}{99^2}\)
\(=\dfrac{1.2.3......98}{2.3.4......99}.\dfrac{3.4.5.....100}{2.3.4.....99}=\dfrac{1}{99}.\dfrac{100}{2}=\dfrac{100}{198}\)
Vậy B = \(\dfrac{100}{198}\)
\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)
Help me, các chế ơi
\(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}=\dfrac{5}{3}\)
`(2^2)/(1 . 3) . (3^2)/(2 . 4) . (4^2)/(3 . 5) . (5^2)/(4 . 6)`
`= 4/3 . 9/8 . 16/15 . 25/24 = 5/3`
\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)
\(X=\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}...\dfrac{99^2}{98.100}\)
\(K=\dfrac{1}{3}.\dfrac{1}{15}.\dfrac{1}{35}...\dfrac{1}{9999}\)
\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
\(\Rightarrow G=\dfrac{2}{3}.\dfrac{96}{505}\)
\(\Rightarrow G=\dfrac{64}{505}\)
\(G=\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}+\dfrac{2}{98.101}\\ G=\dfrac{2}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}+\dfrac{3}{98.101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\\ G=\dfrac{2}{3}.\left(\dfrac{101}{505}-\dfrac{5}{505}\right)\\ G=\dfrac{2}{3}.\dfrac{96}{505}\\ G=\dfrac{64}{505}\)
\(B=\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+...+\frac{99^2}{98.100}\)Tìm phần nguyên của B.
\(B=\left(1+\frac{1}{1.3}\right)+\left(1+\frac{1}{2.4}\right)+\left(1+\frac{1}{3.5}\right)+...+\left(1+\frac{1}{98.100}\right)\)
\(=\left(1+1+1+...+1\right)+\left(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{98.100}\right)\)( 98 số 1 ở tồng đầu tiên)
\(=98+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{97.101}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{98.100}\right)\)
\(=98+\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{3}{97.101}\right)+\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)
\(=98+\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+..+\frac{1}{98}-\frac{1}{100}\right)\)\(=98+\frac{1}{2}.\left(1-\frac{1}{101}\right)+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=98+\frac{1}{2}.\frac{100}{101}+\frac{1}{2}.\frac{49}{100}\)
\(=98+\frac{51}{101}+\frac{49}{200}\)
Suy ra phàn nguyên của B là 98.
Vậy phân fnguyên của B là 98.
mình nhầm. bạn thay các chỗ có "97.101" thành "99.101" nhé!
Chứng tỏ :
a, A = \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2022.2024}\) < \(\dfrac{1}{4}\)
b, B =\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}< \dfrac{1}{2}\)
c, C =\(\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{4}\)
d, D =\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2014^2}< \dfrac{1}{2}\)
a: \(A=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{2022\cdot2024}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2022}-\dfrac{1}{2024}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1011}{2024}=\dfrac{1011}{4848}< \dfrac{1}{4}\)
b: \(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2013\cdot2015}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2013}-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2014}{2015}=\dfrac{1007}{2015}< \dfrac{1}{2}\)
1.Tính giá trị các biểu thức sau
a) \(\left(\dfrac{3}{8}-\dfrac{4}{5}\right).\left(4,34+5,66\right)+\dfrac{1}{4}\)
b) \(\left(\dfrac{3}{2}-\dfrac{5}{6}\right)^2+\left|\dfrac{-5}{2}\right|+\sqrt{\dfrac{4}{9}}\)
c) \(\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+...+\dfrac{99^2}{98.100}\)
2. Cho \(\Delta ABC\) cân tại A. Trên cạnh AB lấy điểm E.Trên tia đối của tia CA lấy điểm F sao cho BE = CF. Nối E với F cắt BC tại O, kẻ EI // AF ( I \(\in\) BC ). Chứng minh rằng: a) \(\Delta\) BEI cân tại E b) OE = OF
Bài 1:
a: \(=\dfrac{15-32}{40}\cdot10+\dfrac{1}{4}\)
\(=\dfrac{-17}{4}+\dfrac{1}{4}=-\dfrac{16}{4}=-4\)
b: \(=\left(\dfrac{9}{6}-\dfrac{5}{6}\right)^2+\dfrac{5}{2}+\dfrac{2}{3}\)
\(=\dfrac{4}{9}+\dfrac{5}{2}+\dfrac{2}{3}\)
\(=\dfrac{8}{18}+\dfrac{45}{18}+\dfrac{12}{18}=\dfrac{65}{18}\)
câu 1 tính tổng \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+....+\dfrac{2}{98.100}\) \(\dfrac{5^2}{1.3}+\dfrac{5^2}{3.5}+...+\dfrac{5^2}{98.100}\) giải hộ mình với ngày mai mình thi rồi
\(\dfrac{2^2}{1.3}.\dfrac{3^3}{2.4}.\dfrac{4^4}{3.5}...\dfrac{50^2}{49.54}\)
với lại lũy thừa tất cả phải là mũ 2
\(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.....\dfrac{50^2}{49.51}\\ =\dfrac{\left(2.3.4.....50\right).\left(2.3.4....50\right)}{\left(1.2.3....49\right).\left(3.4.5.....51\right)}\\ =\dfrac{50.2}{51.1}\\ =1\dfrac{49}{51}\\ =\dfrac{100}{51}\)