Giúp mình nha:
Tính (1/101+ 1/102+.....+1/200): (1/1.2+1/3.4+1/5.6+.....+1/199.200)
Mình cần gấp cảm ơn các bạn !
Bài tập: Thực hiện phép tính sau bằng cách hợp lý nhất
[ 1/1.2 + 1/2.3 + 1/3.4 +........+ 1/199.200 ] : [1/101 + 1/102 + 1/103 +......+ 1/200 ]
Mong các bạn giúp mình giải bai này, mình cảm ơn trước nhé !
Chứng tỏ rằng:
\(\dfrac{\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{199.200}}{\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}}=1\)
Giúp mình với📖
Giúp mình nha:
Bài 1 ; Chứng minh rằng:
a, 1/1.2+1/3.4+5.6 + ........+1/99.100=1/51+1/52+1/53+.....+1/100
b, Tính (1/101+ 1/102+.....+1/200): (1/1.2+1/3.4+1/5.6+.....+1/199.200)
MÌNH CẢM ƠN NHÌU !
Bạn nào giúp mình trước mình tick nha.
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\) (đpcm)
Tính:
.\(\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}\)
Ta có: \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)(1)
Thay (1) vào đề bài
\(\Rightarrow\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}=1\)
\(\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}\)
\(=1\)
Tính: A= (1/101 + 1/102 + 1/103 + ... + 1/200) ÷ (1/1.2 + 1/3.4 + 1/5.6 + ... + 1/ 199.200)
Giúp mình với. Giải chi tiết và đúng mình tik
Bài 1: Tính
a ) D = Cho M = 1/1.2 + 1/3.4 + ........... + 1/199.200
N = 1/101 + 1/102+ ............ + 1/200
Tính M : N
Tính giá trị biểu thức: M=1/1.2+1/3.4+...+1/199.200 chia 1/101+1/102+1/103+...+1/200
cho A =\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\). Biết x<A<y. Tìm x và y.
các bạn ơi giúp mình với tối nay mình cần rồi. Cảm ơn các bạn nhiều.
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+..........+\frac{1}{49.50}\)
\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{49}-\frac{1}{50}\)
\(\Leftrightarrow A=1-\frac{1}{50}=\frac{49}{50}\)
cái kia tự tìm
\(\frac{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...\frac{1}{200}}=1\)
Hãy chứng minh
Đặt \(A=\frac{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}\)
Tử số của A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
\(\Rightarrow A=1\left(đpcm\right)\)