\(\frac{10^3+2\cdot5^5+5^3}{65}\)
Giúp minh với
\(\frac{10^3+2\cdot5^3+5^3}{65}\)
\(\frac{10^3+2.5^3+5^3}{65}=\frac{10^3+5^3.\left(2+1\right)}{65}=\frac{10^3+5^3.3}{65}\)
= \(\frac{10^3+375}{65}=\frac{1375}{65}\)
\(\frac{10^3+2.5^3+5^3}{65}=\frac{1000+2.125+125}{65}=\frac{8.125+2.125+125.1}{65}=\frac{125\left(8+2+1\right)}{65}=\frac{125.11}{65}=\frac{1375}{65}=\frac{275}{13}\)
\(A=\frac{4}{10\cdot2}+\frac{6}{2\cdot20}+\frac{15}{5\cdot20}+\frac{5}{5\cdot40};B=\frac{3}{1\cdot5}+\frac{5}{13\cdot1}+\frac{11}{13\cdot3}+\frac{2}{3\cdot26}\)
So sánh A với B
chứng minh :
\(\frac{4}{3\cdot5}-\frac{6}{5\cdot7}+\frac{8}{7\cdot9}-\frac{10}{9\cdot11}+.......+\frac{2016}{2015\cdot2017}>\frac{1}{6}\)
các bạn giúp nhanh mình cái muộn rồi
chứng minh :
\(\frac{4}{3\cdot5}-\frac{6}{5\cdot7}+\frac{8}{7\cdot9}-\frac{10}{9\cdot11}+.......+\frac{2016}{2015\cdot2017}>\frac{1}{6}\)
các bạn giúp nhanh mình cái muộn rồi
\(\frac{4}{3.5}-\frac{6}{5.7}+\frac{8}{7.9}+\frac{10}{9.11}+...+\frac{2016}{2015.2017}\)
\(=2.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=2.\left(\frac{1}{3}-\frac{1}{2017}\right)\)
\(=2.\frac{2014}{6051}\)
\(=\frac{4028}{6051}\)
\(\Rightarrow BT>\frac{1}{6}\)
Tính nhanh:
\(\frac{\left(\frac{2}{5}\right)^7\cdot5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7\cdot5^2\cdot2^9}\)
Giúp cái nka mấy bn!!!Lm xog giúp mk sẽ chp 5 tik nka!!!Thanks trước...
\(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{9}{4}:\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+\left(\frac{9}{4}.\frac{16}{3}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+3^3.4^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+3^3.2^6}{2^7.5^2.2^9}\)
=\(\frac{2^6.\left(27+2\right)}{2^6.5^2.2^{10}}\)
=\(\frac{29}{25600}\)
Đoạn \(\frac{2^6\cdot\left(27+2\right)}{2^6\cdot5^2\cdot2^{10}}\)là sai rồi bn ơi!!!
Bn phải lm như mục trên là:
\(\frac{2^6\left(2+3^3\right)}{2^7\left(5^2+2^2\right)}\)\(=\frac{2^6\cdot29}{2^7\cdot29}=\frac{1}{2}\)
Nhưng dù sao cx c.ơn bn vì đã giúp mk,mk sẽ cho bn 1 ths nka!!!
bài 1 tính nhanh
a) A=\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
b) B=\(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{57}+...+\frac{3}{49\cdot51}\)
c) C=\(\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}+\frac{5^2}{16\cdot21}+\frac{5^2}{21\cdot26}+\frac{5^2}{26\cdot31}\)
d) D=\(\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+...+\frac{10}{1400}\)
e) E=\(\frac{3}{5\cdot11}+\frac{5}{11\cdot21}+\frac{7}{21\cdot35}+\frac{9}{35\cdot53}\)
f) F=\(\frac{2}{15}+\frac{2}{35}+\frac{2}{99}+\frac{4}{77}\)
giải chi tiết giúp mình nhé thank you very much
A=2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101
A= 2 - 1/3 + 1/3 - 1/5 + 1/5 - ... + 2/99 - 2/101
A = 2 - 2/101 = 200/101
B = 3-1/3+1/3-1/5+1/5-...+3/49-3/51
B = 3-3/51(tự tính nhé)
C = 5(5/1.6+5/6.11+5/11.16+....+5/26-5/31
C = 5(5-1/31)(tự tính)
D rút gon cho 2 rồi 3D , sau đó 5(3/.... tương tự các cách làm trên)
2E nhân lên rồi giải giống trên
3F Rồi nhân 4/77 và rút gọn thì tính được
a, A= \(\frac{1}{1}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+......+\(\frac{1}{99}\)-\(\frac{1}{100}\)
A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+(-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....-\(\frac{1}{99}\)+\(\frac{1}{99}\))
A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+0
A=1-\(\frac{1}{100}\)=\(\frac{100}{100}\)-\(\frac{1}{100}\)=\(\frac{99}{100}\)
a) A= \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}\)
=\(\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right).\frac{3}{2}\)
=\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right).\frac{3}{2}\)
= \(\left(1-\frac{1}{50}\right).\frac{3}{2}=\frac{49}{50}.\frac{3}{2}=\frac{147}{100}\)
c) \(C=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)
= \(\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right).5\)
= \(\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right).5\)
= \(\left(1-\frac{1}{31}\right).5=\frac{30}{31}.5=\frac{150}{31}\)
Mấy bài còn lại mik đang phải nháp đã. Bạn thông cảm cho mik
Cho \(S_1-S_2+S_3-S_4+S_5=\frac{m}{n}\) với m, n nguyên tố cùng nhau. Biết:
\(S_1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(S_2=\frac{1}{2\cdot3}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{2\cdot6}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{3\cdot6}+\frac{1}{4\cdot5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot6}\)
\(S_3=\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot5}+\frac{1}{2\cdot3\cdot6}+\frac{1}{2\cdot4\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot6}+\frac{1}{3\cdot5\cdot6}+\frac{1}{4\cdot5\cdot6}\)
\(S_4=\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot6}+\frac{1}{2\cdot3\cdot5\cdot6}+\frac{1}{2\cdot4\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5\cdot6}\)
\(S_5=\frac{1}{2\cdot3\cdot4\cdot5\cdot6}\)
Tính \(m+n\)
Tính nhanh
\(\frac{4\cdot5\cdot6\cdot7}{14\cdot15\cdot16}\)
\(\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\)
giúp mình với
\(\frac{4.5.6}{14.15.16}\)=\(\frac{1.1.3}{7.3.4}\)=\(\frac{1.1.1}{7.1.4}\)=\(\frac{1}{28}\)
\(\frac{4.5.6.7}{14.15.16.17}=\frac{2.2.5.2.3.7}{2.7.3.5.2.2.2.17}=\frac{1}{2.17}=\frac{1}{34}\) ( Gạch bỏ thừa số giống nhau)
\(\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}=\frac{2.3.4.5}{3.4.5.6}=\frac{1}{3}\)( Gạch bỏ thừa số giống nhau)
Tíc mình nha!
Tính nhanh
\(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{3}{99\cdot100}\)
Giúp mk với T.T
Bạn viết đề sai rồi, mình sửa đề nhé, bài này ngắn lắm =((
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{101}\right)=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}\)(rút gọn phân số)
Ta có :
\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\) ( sai đề rồi )
\(=\)\(\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=\)\(\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\)\(\frac{3}{2}\left(1-\frac{1}{101}\right)\)
\(=\)\(\frac{3}{2}.\frac{100}{101}\)
\(=\)\(\frac{150}{101}\)
Vậy \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}=\frac{150}{101}\)
Chúc bạn học tốt ~