Tính giá trị biểu thức: M=1/1.2+1/3.4+...+1/199.200 chia 1/101+1/102+1/103+...+1/200
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 !
Ta có:1/1.2+1/3.4+1/5.6+...+1/199.200
=1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50
=(1+1/3+1/5+1...199)-2(1/2+1/4+1/6+...+200)
=(1+1/2+1/3+...+1//100)+(1/101+1/102+...+1/200)-(1+1/2+1/3+...+100)
=(1/101+1/102+...+200)=mẫu
bạn xem lại là so sonh hay là tính nha nếu ko minh làm lại cho
mình nhầm chút là cái tử=mẫu thi suy ra tổng đấy =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 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📖
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
\(\Rightarrow\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}=1\)
\(\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}\)
\(=\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}\)
\(=1\)
\(\frac{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...\frac{1}{200}}=1\)
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 = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{199.200}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
\(\Rightarrow A=1\left(đpcm\right)\)
Bài 5 :
a) Chứng minh rằng : 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/199.200/ 1/101 + 1/102 + 1/103 + ... + 1/200 = 1
b) So sánh A = 1 mũ 2/1.2 x 2 mũ 2/2.3 x 3 mũ 2/3.4 x 99 mũ 2/99.100 x 100 mũ 2/100.101 và B = 2 mũ 2/1.3 x 3 mũ 2/2.4 x 4 mũ 2/3.5
x .... x 59 mũ 2/58.60
Chỗ 4 mũ 2/3.5 x ... x 59 mũ 2/58.60 nha
a, Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{199}-\frac{1}{200}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{199}+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
=> \(\frac{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}{\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}}=1\)
=> đpcm
Study well ! >_<
a) Chứng tỏ rằng \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) Đặt A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2013+2014}\); Đặt B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+...+\frac{1}{2014.1008}\)
Chứng tỏ rằng \(\frac{A}{B}\)là số nguyên
