\(\dfrac{7}{1\cdot2}+\dfrac{7}{2\cdot3}+\dfrac{7}{3\cdot4}+...+\dfrac{7}{99\cdot100}\)
giúp mình với mai mình phải nộp rùi!!
tính nhanh.
0.125*\(\dfrac{3}{7}\)-\(\dfrac{1}{8}\)*\(\dfrac{11}{7}\)
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(0,125.\dfrac{3}{7}-\dfrac{1}{8}.\dfrac{11}{7}=\dfrac{1}{8}.\dfrac{3}{7}-\dfrac{1}{8}.\dfrac{11}{7}=\dfrac{1}{8}\left(\dfrac{3}{7}-\dfrac{11}{7}\right)=\dfrac{1}{8}.-\dfrac{8}{7}=-\dfrac{1}{7}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
1)A=\(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+.....+\dfrac{5}{99\cdot100}\)
C=\(1\cdot2\cdot3+2\cdot3\cdot4++3\cdot4\cdot5+4\cdot5\cdot6+5\cdot6\cdot7+6\cdot7\cdot8+7\cdot8\cdot9+8\cdot9\cdot10\)
D=\(1^2+2^2+3^2+...+99^2+100^2\)
a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
\(\dfrac{5}{1\cdot2}+\dfrac{5}{2\cdot3}+\dfrac{5}{3\cdot4}+.....+\dfrac{5}{98\cdot99}+\dfrac{5}{99\cdot100}\)
Giải:
\(\dfrac{5}{1.2}+\dfrac{5}{2.3}+\dfrac{5}{3.4}+...+\dfrac{5}{98.99}+\dfrac{5}{99.100}\)
\(=5.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=5.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=5.\left(1-\dfrac{1}{100}\right)\)
\(=5.\dfrac{99}{100}\)
\(=\dfrac{99}{20}\)
Chúc em học tốt!
Giải:
=5.(11.2+12.3+13.4+...+198.99+199.100)=5.(11.2+12.3+13.4+...+198.99+199.100)
=5.(1−1100)=5.(1−1100)
=9920=9920
Tính tổng:
\(S=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{99\cdot100\cdot101}\)
\(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\)
\(S=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{100.101}\right)\)
\(S=\dfrac{1}{4}-\dfrac{1}{2.100.101}\)
Cho M= \(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+....+\dfrac{1}{99\cdot100}\)
Chứng minh rằng: \(\dfrac{7}{12}< M< \dfrac{5}{6}\)
Tính :
\(A=1\cdot2^2+2\cdot3^2+3\cdot4^2+...+99\cdot100^2\)
Giúp mình nha , chiều mai mình phải học rùi.
thôi đừng giúp nữa , mình nghĩ RA RỒI
Tìm x, biết \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+\left|x+\dfrac{1}{3\cdot4}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=100x\)
Vì \(\left|x+\dfrac{1}{1\cdot2}\right|+\left|x+\dfrac{1}{2\cdot3}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|\ge0\forall x\)
\(\Rightarrow100x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\dfrac{1}{1\cdot2}\right|+...+\left|x+\dfrac{1}{99\cdot100}\right|=x+\dfrac{1}{1\cdot2}+...+x+\dfrac{1}{99\cdot100}\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\dfrac{1}{1\cdot2}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow99x+\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)=100x\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}=x\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=x\)
\(\Rightarrow x=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(E=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{98\cdot99\cdot100}\)
\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
\(E=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
\(E=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{4949}{9900}\)
\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{98.99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right)\)
...
E = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
2E = \(\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}\)
= \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}\right)\)+\(\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\) +...+ \(\left(\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
= \(\dfrac{1}{1.2}\) - \(\dfrac{1}{99.100}\)= \(\dfrac{1}{2}-\dfrac{1}{9900}\) = \(\dfrac{4949}{9900}\)
tính tổng
S=\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\)
A=1+\(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)
B= \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...\dfrac{1}{100}\)
C=\(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
a) Ta có
S = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)2S = \(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)
S = \(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right).\left(n+2\right):2}\)
b) A = \(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)
A = \(2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A = \(2-\dfrac{1}{99}\)
A = \(\dfrac{197}{99}\)
c) Ta có
B = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
B = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B = \(1-\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
d) Ta có
C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
C = \(1+\left(1+\dfrac{98}{2}\right)+\left(1+\dfrac{97}{3}\right)+...+\left(1+\dfrac{1}{99}\right)\)
C = \(1+50+\dfrac{100}{3}+...+\dfrac{100}{99}\)
C = 51 + 100(\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\))
Đặt D = \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{99}\)
D = \(\dfrac{97}{198}\)
=> C = 51 + 100.\(\dfrac{97}{198}\)
C = 51 + \(\dfrac{4850}{99}\)
C = \(\dfrac{9899}{99}\)
Đây là bài làm của mình sai thì nx nha