so sánh các phân số sau (ko quy đồng mẫu)
A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\) với 1
trình bày rõ ràng
so sánh các phân số sau (ko quy đồng mẫu)
A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\) với 1
trình bày rõ ràng
Ta có : \(\frac{1}{2.3}< \frac{1}{1.2}\)
\(\frac{1}{3.4}< \frac{1}{2.3}\)
\(\frac{1}{4.5}< \frac{1}{3.4}\)
...
\(\frac{1}{99.100}< \frac{1}{98.99}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\)
\(A< 1-\frac{1}{99}< 1\)
\(\Rightarrow A< 1\)
A \(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Vì \(\frac{49}{100}< 1\Rightarrow A< 1\)
Chúc bn hk tốt :>
mẹt xi
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 < \(\frac{1}{2}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(\Rightarrow A< 1\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow B< \frac{1}{2}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}< 1\)
\(B=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}=?\)
Làm bậy, mà đúng
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ … + \(\frac{1}{99.100}\)
= \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ … + \(\frac{1}{99}\)- \(\frac{1}{100}\)
= \(\frac{1}{1}\)- \(\frac{1}{100}\)
= \(\frac{99}{100}\)
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
Cho \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
So sánh A với 1
A=1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1-1/100
SUY RA A<1 VÌ 1/100>0
Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=1-\frac{1}{100}\)
\(\Rightarrow A=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\Rightarrow A< 1\)
Vậy A < 1
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{100.101}\)
So sánh A và B
Các bạn vào đây : ~ Marshmallow ~
Hoặc : https://olm.vn/thanhvien/cutedangyeu___
Đọc stt nhé, cảm ơn!
Tính: \(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
(Tính nhanh, ghi rõ cách tính nha các bạn)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\) . Tính
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}\)
\(=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+....+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Tính A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+......+\frac{1}{99.100}=?\)
mk bít lm cách lớp 5, vừa học
Cần ko bn
Bài: tính
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
= 1/2-1/3+ 1/3 -1/4 +... +1/99-1/100
=1/2-1/100
=50/100 - 1/100= 49/100
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Tham khảo nha !!!
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{50}{100}-\frac{1}{100}\)
\(=\frac{49}{100}\)