Chứng minh rằng
\(A=\dfrac{8}{1^2.3^2}+\dfrac{16}{3^2.5^2}+\dfrac{24}{5^2.7^2}+...+\dfrac{120}{29^2.31^2}< 1\)
Tính:
a) S=1.2+2.3+3.4+...+99.100
b) B=\(\dfrac{49^{24}.125^{17}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)
c) C=\(\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}\right).3^5+\left(\dfrac{1}{3^5}+\dfrac{1}{3^6}+\dfrac{1}{3^7}+\dfrac{1}{3^8}\right).3^9+...+\left(\dfrac{1}{3^{97}}+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\right).3^{101}\)
d) D= \(3-3^2+3^3-3^4+...+3^{2017}-3^{2018}\)
a) S = 1.2 + 2.3 + 3.4 + ... + 99.100
S có thể được viết lại thành:
S = 1(2 - 0) + 2(3 - 1) + 3(4 - 2) + ... + 99(100 - 98)
= 1.2 - 0 + 2.3 - 1 + 3.4 - 2 + ... + 99.100 - 98
= (1.2 + 2.3 + 3.4 + ... + 99.100) - (0 + 1 + 2 + ... + 98)
Để tính tổng 1.2 + 2.3 + 3.4 + ... + 99.100, ta sử dụng công thức:
S = n(n+1)(2n+1)/6
Với n = 99, ta có:
S = 99.100.199/6 = 331650
Tính tổng 0 + 1 + 2 + ... + 98, ta sử dụng công thức:
S = n(n+1)/2
Với n = 98, ta có:
S = 98.99/2 = 4851
Do đó, S = 331650 - 4851 = 326799
b) B = 4924.12517.28−530.749.45529.162.748
B có thể được viết lại thành:
B = (4924.12517.28) / (530.749.45529.162.748)
B = (4924 / 530) . (12517 / 749) . (28 / 45529) . (162 / 162) . (748 / 748)
B = 9.17.28/45529 = 2^2 . 3^2 . 17 / 45529
B = 108 / 45529
c) C = (13+132+133+134).35+(135+136+137+138).39+...+(1397+1398+1399+13100).3101
C = (13(1 + 13 + 13^2 + 13^3)) . 3^5 + (13^5(1 + 13 + 13^2 + 13^3)) . 3^9 + ... + (13^97(1 + 13 + 13^2 + 13^3)) . 3^101
C = (1 + 13 + 13^2 + 13^3) . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)
C = 80 . (13^5 . 3^5 + 13^9 . 3^9 + ... + 13^97 . 3^101)
C = 80 . (13^5 . 3^4 . 3 + 13^9 . 3^8 . 3 + ... + 13^97 . 3^96 . 3)
C = 80 . (13^6 . 3^5 + 13^10 . 3^9 + ... + 13^98 . 3^97)
C = 80 . 3^5 (13^6 + 13^10 + ... + 13^98)
d) D = 3 - 3^2 + 3^3 - 3^4 + ... + 3^2017 - 3^2018
D = (3 - 3^2) + (3^3 - 3^4) + ... + (3^
Tính :
a) \(\dfrac{17}{23}.\dfrac{8}{16}.\dfrac{23}{17}.\left(-80\right).\dfrac{3}{4}\)
b) \(\dfrac{5}{11}.\dfrac{18}{29}-\dfrac{5}{11}.\dfrac{8}{29}+\dfrac{5}{11}.\dfrac{19}{29}\)
c) \(\left(\dfrac{13}{23}+\dfrac{1313}{2323}-\dfrac{131313}{232323}\right).\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{7}{12}\right)\)
d) \(\dfrac{1^2}{1.2}.\dfrac{2^{2^{ }}}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}.\dfrac{5^2}{5.6}.\dfrac{6^2}{6.7}.\dfrac{7^2}{7.8}.\dfrac{8^2}{8.9}.\dfrac{9^2}{9.10}\)
e) \(\dfrac{2^2}{3}.\dfrac{3^2}{8}.\dfrac{4^2}{15}.\dfrac{5^2}{24}.\dfrac{6^2}{35}\dfrac{7^2}{48}.\dfrac{8^2}{63}.\dfrac{9^2}{80}\)
đây là tính nhanh à nếu tính bình thường thì tính may tính là ra
a) 17/23 . 8/16 . 23/17. (-80) . 3/4
= (17/23 . 23/17) . (8/16 . 3/4) . (-80)
= 1 . 3/8 . (-80)
= 3/8 . (-80)
= -30
b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29
= 5/11 . (18/29 - 8/29 + 19/29)
= 5/11 . 1
= 5/11
c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)
= (13/23 + 1313/2323 - 131313/232323).0
= 0
d) 12/2x2 . 22/2x3 . 32/3x4 . 42/4x5 . 52/5x6 . 62/6x7 . 72/7x8 . 82/8x9 . 92/9x10
= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10
= 1/10
Khó nhìn quá. Bạn thông cảm nhé!
Các bạn giúp với :<
Bài 1:
a, CMR: A = \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{21}{10^2.11^2}< 1\)
b, Cho B = \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}.\) CMR: B không phải là số nguyên.
c, So sánh: C = \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2021}{2^{2020}}\) với 3.
Chứng minh rằng :
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\)
Chứng minh rằng :
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}< 1\)
chứng minh rằng
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
Chứng minh rằng : \(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}\)
\(A=\dfrac{2^2}{1^2.2^2}-\dfrac{1^2}{1^2.2^2}+\dfrac{3^2}{2^2.3^2}-\dfrac{2^2}{2^2.3^2}+...+\dfrac{10^2}{9^2.10^2}-\dfrac{9^2}{9^2.10^2}\)\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
\(A=1-\dfrac{1}{10^2}< 1\left(đpcm\right)\)
A=\(\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+....+\dfrac{1}{9^2}-\dfrac{1}{10^2}\)
A = \(\dfrac{1}{1^2}-\dfrac{1}{10^2}\)
A = \(1-\dfrac{1}{10^2}\) < 1
Vậy A < 1
3. b) \(1\dfrac{13}{15}\).(0,5)2.3+(\(\dfrac{8}{15}\)-\(1\dfrac{19}{60}\)):\(1\dfrac{23}{24}\)
c) (-2)3.\(\dfrac{-1}{24}\)+(\(\dfrac{4}{5}\)-1,2):\(\dfrac{2}{15}\)
d) (\(\dfrac{-2}{5}\))2+\(\dfrac{1}{2}\).(4,5-2)-255
Bài 1:
a, Tính \(\dfrac{5.4^{15}.9^{9} - 4.3^{20}.8^{9}}{5.2^{9}.6^{19}-7.2^{29}.27^{6}}\)
b,Tìm x biết:
\(1\dfrac{1}{30}:(24\dfrac{1}{6}-24\dfrac{1}{5}) -\dfrac{1\dfrac{1}{2}-\dfrac{3}{4}}{4x-\dfrac{1}{2}}=(-1\dfrac{1}{15}):(8\dfrac{1}{5}-8\dfrac{1}{3})\)
Bài 2: Chứng minh số:
222...22200333...333(2001c/s 2; 2003 c/s3)