a, xét từng đoạn 1 , 1/2 ,1/2^3 ,1/2^4 ,1/2^5 ,1/2^6
ta có
1 = 1
1/2 + 1/3 < 1/2 + 1/2 = 1
1/4 + 1/5 + .. + 1/7 < 1/4 +..+ 1/4 = 4/4 = 1
1/8 + 1/9 + .. + 1/15 < 1/8 + .. + 1/8 = 8/8 = 1
tương tự
1/16 +1/17 + .. + 1/31 < 1
1/32 + 1/33 + .. + 1/63 < 1
=> cộng lại => A < 6
b, Câu hỏi của trịnh quỳnh trang - Toán lớp 6 - Học toán với OnlineMath
bài 4 : cmr :
c) C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{9999}{10000}< \dfrac{1}{100}\)
CMR 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 \(\dfrac{1}{100}\)
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: \(\left[\left(1+2+3+....+n\right)-7\right]⋮̸10\)
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.
Chứng minh rằng :
\(\dfrac{1}{6}< \dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{100^2}< \dfrac{1}{4}\)
Chứng minh rằng : \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)
chứng tỏ B = \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{63}< 6\)
1 thực hiện phép tính
a,\(\dfrac{-2}{3}+\dfrac{3}{4}-\dfrac{-1}{6}+\dfrac{-2}{6}-\dfrac{-2}{5}\)
b,\(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{-7}{10}\)
c,\(\dfrac{1}{2}-\dfrac{-2}{5}+\dfrac{1}{3}+\dfrac{5}{7}-\dfrac{-1}{6}+\dfrac{-4}{35}+\dfrac{1}{41}\)
d,\(\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
