Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,4^{10}\cdot8^{15}=2^{20}\cdot2^{45}=2^{65}\)
\(b,4^{15}\cdot5^{30}=4^{15}\cdot5^{15}\cdot5^{15}=\left(4\cdot5\cdot5\right)^{15}=100^{15}\)
\(c,27^{16}\cdot9^{10}=3^{48}\cdot3^{20}=3^{68}\)
a) \(4^{10}.8^{15}=\left(2^2\right)^{10}.\left(2^3\right)^{15}=2^{20}.2^{45}=2^{20+45}=2^{65}\)
b) \(4^{15}.5^{30}=\left(2^2\right)^{15}.5^{30}=2^{30}.5^{30}=\left(2.5\right)^{30}=10^{30}\)
c) \(27^{16}.9^{10}=\left(3^3\right)^{16}.\left(3^2\right)^{10}=3^{48}.3^{20}=3^{48+20}=3^{68}\)
a: \(=\dfrac{2^{12}\cdot3^{14}}{3^{12}\cdot2^{12}}=3^2=9\)
b: \(=\dfrac{7^3\cdot2\cdot5^3}{5^2\cdot7^2\cdot6}=7\cdot5\cdot\dfrac{1}{3}=\dfrac{35}{3}\)
d: =2^5(2^8+1)/2^2(2^8+1)=2^3=8
c: \(=\dfrac{5^3\cdot3^6\cdot2^8\cdot5^4\cdot3^4\cdot2^2}{2^{10}\cdot3^{10}\cdot5^5}=5^2\)
\(M=\dfrac{2^2.3^2.4^2.....20^2}{1.3.2.4.3.5.4.6.5.7.6.8.7.9....19.21}=\)
\(=\dfrac{2^2.3^2.4^2....20^2}{1.2.3^2.4^2....19^2.20.21}=\dfrac{2.20}{21}=\dfrac{40}{21}\)
\(N=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{10}{11}=\dfrac{1}{11}\)
\(=1+1+1+1+1+1+1+1+1+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=9+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
=9+9/10=99/10
a)\(3-\left(\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
=\(3-\frac{1}{4}-\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
=\(\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{3}-\frac{1}{3}\right)+\left(\frac{6}{5}-\frac{3}{2}\right)\)
=\(-8+\frac{3}{2}-1-\frac{3}{10}\)
=\(\left(-8-1\right)+\left(\frac{3}{2}-\frac{3}{10}\right)\)
=-9+\(\frac{6}{5}\)
=\(\frac{-39}{5}\)
a)
1/12 + 1/6 + 1/2 = (1+2+6)/12 = 9/12 = 3/4
1/30 + 1/20 = (2+3)/60 = 5/60 = 1/12
1/56 + 1/42 = 1/7(1/8+1/6) = 1/7 .(3+4)/24 = 1/24
8/9- 1/72 = (8.8 - 1)/72 = 63/72 = 7/8
1/12 + 1/24 = (2+1)/24 = 1/8
7/8 - 1/8 = 6/8 = 3/4
3/4 - 3/4 = 0
b)
\(0,5+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{6}\right)+\left(\frac{5}{7}-\frac{4}{35}\right)\)
\(=\frac{15+10+12+5}{30}+\frac{25-4}{35}\)
\(=\frac{7}{5}+\frac{3}{5}\)
\(=2\)
1. \(\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\frac{3^{10}\left(11+5\right)}{3^9\cdot2^4}=\frac{3^{10}\cdot2^4}{3^9\cdot2^4}=3\)
2. \(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\frac{2^{10}\cdot\left(13+65\right)}{2^8\cdot104}=\frac{2^{10}\cdot78}{2^8\cdot104}=\frac{2^8\cdot2^2\cdot2\cdot3\cdot13}{2^8\cdot2^3\cdot13}=\frac{2^8\cdot2^3\cdot3\cdot13}{2^8\cdot2^3\cdot13}=3\)
3. \(\frac{72^2\cdot54^2}{108^4}=\frac{\left(2^3\cdot3^2\right)^2\cdot\left(2\cdot3^3\right)^2}{\left(2^2\cdot3^3\right)^4}\)
\(=\frac{2^6\cdot3^4\cdot2^2\cdot3^6}{2^8\cdot3^{12}}=\frac{2^8\cdot3^{10}}{2^8\cdot3^{12}}=\frac{3^{10}}{3^{12}}=3^{-2}=\frac{1}{9}\)
4. \(\frac{21^2\cdot14\cdot125}{35^5\cdot6}=\frac{\left(3\cdot7\right)^2\cdot2\cdot7\cdot5^3}{\left(5\cdot7\right)^5\cdot2\cdot3}=\frac{3^2\cdot7^2\cdot2\cdot7\cdot5^3}{5^5\cdot7^5\cdot2\cdot3}=\frac{3^2\cdot7^3\cdot2\cdot5^3}{5^3\cdot5^2\cdot7^2\cdot7^3\cdot2\cdot3}=\frac{3^2}{5^2\cdot3\cdot7^2}=\frac{3}{1225}\)