Các bạn giải giúp mình nhé! Thank you !
A = \(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+...+\frac{4}{11\cdot13}\)
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
giúp mk nha các bn
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
\(\frac{\frac{3}{5}\cdot7^2-3,5^6+\frac{3}{5}\cdot3^9}{\frac{3}{4}\cdot7^2-\frac{3}{4}\cdot5^7+\frac{3}{4}\cdot3^9}\)
\(\frac{2^5\cdot7+2^5}{2^5\cdot2^5-2^5\cdot3}\)và \(\frac{3^4\cdot5\cdot< -3>^6}{3^4\cdot13\cdot3^4}\)
Quy Đồng gium mk nha.truoc 2 h voi help help
ta có: \(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(2^5-3\right)}=\frac{8}{2^5-3}=\frac{8}{29}=\frac{104}{377}\)
\(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}=\frac{3^{10}.5}{3^8.13}=\frac{3^2.5}{13}=\frac{45}{13}=\frac{1305}{377}\)
\(\Rightarrow\frac{104}{377}< \frac{1305}{377}\Rightarrow\frac{2^5.7+2^5}{2^5.2^5-2^5.3}< \frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)
Ta cứ tính ra tử số và mỗi số của từng phân số ra nhé Jerry Gaming:
\(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}\)= \(\frac{2^5.\left(7+1\right)}{2^5.\left(2^5-3\right)}=\frac{2^5.8}{2^5.\left(32-3\right)}=\frac{32.8}{2^5.29}=\frac{32.8}{32.29}=\frac{8}{29}\)
\(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)= \(\frac{3^4.5.3^6}{3^8.13}=\frac{3^{10}.5}{3^8.13}=\frac{3^2.5}{13}=\frac{9.5}{13}=\frac{45}{13}\)
\(\frac{8}{29}\)và \(\frac{45}{13}\)MSC: 377
Ta có:
\(\frac{8}{29}=\frac{8.13}{29.13}=\frac{104}{377}\)
\(\frac{45}{13}=\frac{45.29}{13.29}=\frac{1305}{377}\)
Vậy quy đồng \(\frac{2^5.7+2^5}{2^5.2^5-2^5.3}\)và \(\frac{3^4.5.\left(-3\right)^6}{3^4.13.3^4}\)ta được \(\frac{104}{377}\)và \(\frac{1305}{377}\)
Chúc bạn học tốt!
bài 1 tính tổng
a) \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
b) \(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}\)
bài 2 chứng tỏ rằng phân số \(\frac{2n+1}{3n+2}\)là phân số tối giản.
bài 3 cho A=\(\frac{n+2}{n-5}\)(n thuộc z;n khác 5) tìm x để A thuộc z
bài 4 tính giá trị biểu thức
A=\(10101\cdot\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3\cdot7\cdot11\cdot13\cdot37}\right)\)
Bài 1 :
a) =) \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)= \(1-\frac{1}{101}=\frac{100}{101}\)
b) =) \(\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
=) \(\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)( theo phần a)
Bài 2 :
-Gọi d là UCLN \(\left(2n+1;3n+2\right)\)( d \(\in N\)* )
(=) \(2n+1⋮d\left(=\right)3.\left(2n+1\right)⋮d\)
(=) \(6n+3⋮d\)
và \(3n+2⋮d\left(=\right)2.\left(3n+2\right)⋮d\)
(=) \(6n+4⋮d\)
(=) \(\left(6n+4\right)-\left(6n+3\right)⋮d\)
(=) \(6n+4-6n-3⋮d\)
(=) \(1⋮d\left(=\right)d\in UC\left(1\right)\)(=) d = { 1;-1}
Vì d là UCLN\(\left(2n+1;3n+2\right)\)(=) \(d=1\)(=) \(\frac{2n+1}{3n+2}\)là phân số tối giản ( đpcm )
Bài 3 :
-Để A \(\in Z\)(=) \(n+2⋮n-5\)
Vì \(n-5⋮n-5\)
(=) \(\left(n+2\right)-\left(n-5\right)⋮n-5\)
(=) \(n+2-n+5⋮n-5\)
(=) \(7⋮n-5\)(=) \(n-5\in UC\left(7\right)\)= { 1;-1;7;-7}
(=) n = { 6;4;12;-2}
Vậy n = {6;4;12;-2} thì A \(\in Z\)
Bài 4:
A = \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
= \(10101.\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{111111}\right)\)
= \(10101.\left(\frac{1}{111111}+\frac{5}{222222}\right)\)= \(10101.\left(\frac{2}{222222}+\frac{5}{222222}\right)\)
= \(10101.\frac{7}{222222}\)( không cần rút gọn \(\frac{7}{222222}\))
= \(\frac{7}{22}\)
tính rồi so sánh m và n biết
m = \(\frac{3}{1\cdot3}\)+ \(\frac{3}{3\cdot5}\)+\(\frac{3}{5\cdot7}\)+...+\(\frac{2}{99\cdot100}\)
n = \(\frac{3}{1\cdot3}\)+ \(\frac{3}{3\cdot5}\)+ \(\frac{3}{5\cdot7}\)+ ...+ \(\frac{3}{97\cdot99}\)
rồi, mấy bạn giải ra giùm mk nhé. 4h mk di học rồi cho nên trước 4 giờ các bạn giải cho mk nhé.
n=\(\frac{2}{3}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\left(1-\frac{1}{99}\right)\)
n=\(\frac{2}{3}\times\frac{98}{99}\)
n=\(\frac{196}{297}\)
Câu \(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{2}{99.100}\)Bạn viết \(\frac{3}{99.100}=\frac{2}{99.100}\)mik sửa lại nhé.
\(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.100}\)
\(M=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{100-99}{99.100}\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)\)
\(M=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)
\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{97.99}\)
\(N=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{99-97}{97.99}\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)\)
\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{3}{2}.\frac{98}{99}=\frac{49}{33}\)
Ta thấy : \(\frac{297}{200}>\frac{49}{33}\Rightarrow M>N\)
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
Bài 1: Rút gọn rồi quy đồng
\(\frac{4\cdot5+4\cdot11}{8\cdot7-4\cdot3}\) \(\frac{-15\cdot8+10\cdot7}{5\cdot6+20\cdot3}\)và \(\frac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^3\cdot11}\)
\(\frac{16}{11},-\frac{5}{9},\frac{10}{539}\)
\(\frac{1}{1\cdot3\cdot7}\)+\(\frac{1}{3\cdot7\cdot9}\)+\(\frac{1}{7\cdot9\cdot13}\)+\(\frac{1}{9\cdot13\cdot15}\)+\(\frac{1}{13\cdot15\cdot19}\)
Giải giúp mình nhanh lên nhé
\(\frac{1}{1.3.7}+\frac{1}{3.7.9}+\frac{1}{7.9.13}+\frac{1}{9.13.15}+\frac{1}{13.15.19}\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.3}-\frac{1}{15.19}\right)=\frac{47}{285}\)
Tính nhanh biểu thức sau :
\(\frac{4}{1\cdot3}+\frac{4}{3\cdot5}+\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+....+\frac{4}{15\cdot17}+\frac{4}{17\cdot19}+\frac{4}{19\cdot21}\)
\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{19.21}\right)\)
=\(2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
=\(2\left(1-\frac{1}{21}\right)\)
=\(\frac{2.20}{21}=\frac{40}{21}\)