Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: -8/31=-808/3131
-786/3131=-786/3131
b: \(\dfrac{11}{2^3\cdot3^4\cdot5^2}=\dfrac{11\cdot5}{2^3\cdot3^4\cdot5^3}=\dfrac{55}{2^3\cdot3^4\cdot5^3}\)
\(\dfrac{29}{2^2\cdot3^4\cdot5^3}=\dfrac{29\cdot2}{2^3\cdot3^4\cdot5^3}=\dfrac{58}{2^3\cdot3^4\cdot5^3}\)
c: 7/39=140/780
11/65=132/780
9/52=135/780
Quy đồng các phân số sau
a)−8/31;−789/3131
b)11/2^3.3^4.5^2;29/2^2.2^4.5^3
c)1/n và 1/n+1 (n thuộc N)
a) \(\dfrac{-8}{31}=\dfrac{-8\cdot101}{31\cdot101}=\dfrac{-808}{3131}\)
\(\dfrac{-789}{3131}=\dfrac{-789}{3131}\)
c) \(\dfrac{1}{n}=\dfrac{n+1}{n\left(n+1\right)}\)
\(\dfrac{1}{n+1}=\dfrac{n}{n\left(n+1\right)}\)
Quy đồng các phân số sau
a)−8/31;−789/3131
b)11/2^3.3^4.5^2;29/2^2.2^4.5^3
c)1/n và 1/n+1 (n thuộc N)
a) \(-\dfrac{8}{31}=\dfrac{-8\cdot101}{31\cdot101}=\dfrac{-808}{3131}\)
\(\dfrac{-789}{3131}=\dfrac{-789}{3131}\)
c) \(\dfrac{1}{n}=\dfrac{n+1}{n\left(n+1\right)}\)
\(\dfrac{1}{n+1}=\dfrac{n}{n\left(n+1\right)}\)
a: \(\dfrac{4\cdot5+4\cdot11}{8\cdot7+4\cdot3}=\dfrac{20+44}{56+12}=\dfrac{64}{68}=\dfrac{16}{17}=\dfrac{11088}{11781}\)
\(\dfrac{-15\cdot8+10\cdot7}{5\cdot6+20\cdot3}=\dfrac{-120+70}{30+60}=\dfrac{-50}{90}=\dfrac{-5}{9}=\dfrac{-6545}{11781}\)
\(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=\dfrac{2\cdot5}{7\cdot11}=\dfrac{10}{77}=\dfrac{1530}{11781}\)
Câu 1:
\(\dfrac{14\cdot34-21\cdot10}{28\cdot10-28\cdot16}=\dfrac{7\cdot2\cdot34-7\cdot3\cdot10}{28\left(10-16\right)}\)
\(=\dfrac{7\left(2\cdot34-3\cdot10\right)}{28\cdot\left(-6\right)}=\dfrac{1}{4}\cdot\dfrac{68-30}{-6}=\dfrac{1}{4}\cdot\dfrac{38}{-6}=\dfrac{38}{-24}=-\dfrac{19}{12}\)
a) \(-\dfrac{8}{31}=\dfrac{-8\cdot101}{31\cdot101}=\dfrac{-808}{3131}\)
\(\dfrac{-789}{3131}=\dfrac{-789}{3131}\)
b) \(\dfrac{11}{2^3\cdot3^4\cdot4^5}=\dfrac{11\cdot2^3\cdot5^3}{2^6\cdot3^4\cdot4^5\cdot5^3}=\dfrac{11000}{2^6\cdot3^4\cdot4^5\cdot5^3}\)
\(\dfrac{29}{2^2\cdot2^4\cdot5^3}=\dfrac{29\cdot3^4\cdot4^5}{2^6\cdot3^4\cdot4^5\cdot5^3}=\dfrac{2405376}{2^6\cdot3^4\cdot4^5\cdot5^3}\)
c) \(\dfrac{1}{n}=\dfrac{n+1}{n\left(n+1\right)}\)
\(\dfrac{1}{n+1}=\dfrac{n}{n\left(n+1\right)}\)