Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
giúp mk nha. mk sẽ k cho bn nào trả lời giúp mk mà đúng
(1/1*2+1/2*3+1/3*4+...+1/8*9+1/9*10)*100-[5/2:(x+206/100)]:1/2=89
Đặt A=1/1*2+1/2*3+1/3*4+...+1/8*9+1/9*10
A=1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9+1/9-1/10
A=1-1/10
A=9/10
=>(1/1*2+1/2*3+1/3*4+...+1/8*9+1/9*10)*100-[5/2:(x+206/100)]:1/2=89
=9/10*100-[5/2:(x+206/100)]:1/2=89
90-[5/2:(x+206/100)]:1/2=89
5/2:(x+206/100):1/2=90-89
5/2:(x+206/100):1/2=1
x+206/100:1/2=5/2:1
x+206/100:1/2=5/2
x+103/25=5/2
x=5/2-103/25
x=-81/50
\(A=\frac{1}{2.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\frac{1}{4}+\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)\)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)
( gạch bỏ các phân số giống nhau)
\(A=\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(A=\frac{1}{4}+\frac{2}{9}\)
\(A=\frac{17}{36}\)
phần b, c bn lm tương tự như phần a nha
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
= \(1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{98}-\frac{1}{98}\right)-\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
= \(1-\frac{1}{100}\)
= \(\frac{99}{100}\)
Vậy ...
B = \(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{17.20}\)
= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
= \(\frac{1}{2}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{17}-\frac{1}{17}\right)-\frac{1}{20}\)
= \(\frac{1}{2}-\frac{1}{20}\)
= \(\frac{9}{20}\)
Vậy B = 9/20
\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)
\(=\frac{1.1.2.2.3.3...100.100}{1.2.2.3.3.4.4...100.101}\)
\(=\frac{\left(1.2.3...100\right)\left(1.2.3...100\right)}{\left(1.2.3..100\right)\left(2.3.4...101\right)}=\frac{1}{101}\)
A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
Chứng tỏ rằng;\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+......+\(\frac{1}{49.50}\) < 1
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1-\frac{1}{50}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)\)
\(=\left(1-\frac{1}{50}\right)+0+0+...+0=1-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}\)
Vì \(\frac{49}{50}<1\) nên \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}<1\)
Vậy thỏa mãn đề bài
Ta có:
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
Mà \(\frac{49}{50}< 1\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}< 1\left(đpcm\right)\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)\(=\frac{9}{10}\)