a) Rút gọn: \(\frac{9.25-63}{3.30+153}\)
b) Chứng tỏ rằng:
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2}\)
Những câu hỏi liên quan
a) Rút gọn: \(\frac{9.25-63}{3.30+153}\)b) Chứng tỏ rằng: \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}
Xem chi tiết
Tổng quát: \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\) (với mọi số tự nhiên n khác 0)
Ta có: \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}<\frac{1}{2}\) (vì \(\frac{1}{100}>0\) )
=>đpcm
Đúng 0
Bình luận (0)
\(\frac{9.25-63}{9.10+153}\)=\(\frac{9.25-9.7}{9.10+9.17}\)=\(\frac{9.\left(25-7\right)}{9.\left(10+17\right)}\)=\(\frac{9.18}{9.27}\)=\(\frac{1.2}{1.3}\)=\(\frac{2}{3}\)
Đúng 0
Bình luận (0)
chứng tỏ rằng: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)
1/1.2 + 1/2.3 + 1/3.4 + .......................+ 1/99.100
= 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 +..................+ 1/99 - 1/100
= 1 - 1/100
= 99/100
Đúng 0
Bình luận (0)
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
Ma 99/100 < 1.
=> 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 < 1 (dccm)
Đúng 0
Bình luận (0)
Chứng tỏ rằng
\(a,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)
\(b,\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
A.\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
So sánh A với 1
B.\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
So sánh B với \(\frac{1}{2}\)
A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=\(1-\frac{1}{50}\)
Vì \(1-\frac{1}{50}< 1\)nên A < 1
B = \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{2}-\frac{1}{100}\)
Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)
Đúng 0
Bình luận (0)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(\Rightarrow A< 1\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow B< \frac{1}{2}\)
Đúng 0
Bình luận (0)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}< 1\)
\(B=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
tính giá trị của biểu thức
a) A=\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\) + ...+\(\frac{1}{99.100}\)
b) B= \(\frac{2}{1.3}\)+\(\frac{2}{3.5}\) + \(\frac{2}{5.7}\)+\(\frac{2}{7.9}\) +...+\(\frac{2}{97.99}\)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
Đúng 0
Bình luận (1)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
Đúng 0
Bình luận (3)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
=>\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=>\(\frac{1}{1}-\frac{1}{100}\)
=>\(\frac{99}{100}\)
B=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\)
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)
=>\(\frac{1}{1}-\frac{1}{99}\)
=>\(\frac{98}{99}\)
Đúng 0
Bình luận (0)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=?\)
Làm bậy, mà đúng
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Đúng 0
Bình luận (0)
\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ … + \(\frac{1}{99.100}\)
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ … + \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(\frac{1}{1}\)- \(\frac{1}{100}\)
= \(\frac{99}{100}\)
Đúng 0
Bình luận (0)
1/1 . 2 + 1/ 2 . 3 + 1/ 3 . 4 + ... + 1/99 . 100
= 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1/1 - 1/100
= 100/100 + -1/100
= 99/100
#Hoq chắc _ Baccanngon
Đúng 0
Bình luận (0)
Chứng tỏ rằng:\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}< 1\)
Ai nhanh nhất sẽ đc tick
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
Đúng 0
Bình luận (0)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\)
Đúng 0
Bình luận (0)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Chứng minh rằng:
a)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}< \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
b)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}< 1-\frac{1}{2.3}\)
Cần gấp, ai nhanh mik tick nha
Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn
\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)
\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)
\(B=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)....\left(1-\frac{2}{99.100}\right)\)