ta có :
1/2 < 2/3
2/3 <3/4
.........
9999/10000 < 10000/10001
suy ra : A2 < 1/22/33/4*****9999/1000010000/10001
suy ra : A2 < 1/10001 < 1/10000= (1/100)2
suy ra A2 < (1/100)2 . Từ đó: A < 1/100
2 là mũ 2 nha bạn
bài 4 : cmr :
c) C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{9999}{10000}< \dfrac{1}{100}\)
bài 5
cho A=\(\dfrac{1}{2}\cdot\dfrac{3}{4}\dfrac{5}{6}\cdot...\cdot\dfrac{9999}{10000}\)
So sánh a với \(\dfrac{1}{100}\)
Chứng minh rằng:
a, A= 1+\(\dfrac{1}{2}\)+ \(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+......+\(\dfrac{1}{63}\)<6
b, B= \(\dfrac{1}{2}\).\(\dfrac{3}{4}\).\(\dfrac{5}{6}\)......\(\dfrac{9999}{10000}\)< \(\dfrac{1}{100}\)
Câu 1: Tìm x biết
a) \(-\dfrac{2}{3}\)\(\left(x-\dfrac{1}{4}\right)\) = \(\dfrac{1}{3}\left(2x-1\right)\) b) \(\dfrac{1}{5}.2^x+\dfrac{1}{3}.2^{x+1}=\dfrac{1}{5}.2^7+\dfrac{1}{3}.2^8\)
Câu 2: a) Cho A = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}......\dfrac{9999}{10000}\)
So sánh A vs 0,01
b) Chứng tỏ rằng: \(\left[\left(1+2+3+....+n\right)-7\right]⋮̸10\)
K=\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)
CMR:\(\dfrac{1}{5}< K< \dfrac{1}{3}\)
CMR:
a) \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
b) \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}>48\)
Chứng minh rằng: \(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}>98\)
Cho C = \(\dfrac{3}{4}\) +\(\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}\)
Chứng minh rằng C>98
Chứng minh rằng:
1) B =\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}>1\)
2) \(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+\dfrac{4}{5^4}+...+\dfrac{500}{5^{500}}<100\)
3) \(C=\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{500^3}<\dfrac{1}{4}\)
4) \(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}<100\)
Làm giúp mình sớm nha! Thanks.