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b) \(\frac{36^7}{2^{15}.27^5}=\frac{\left(2^2.3^2\right)^7}{2^{15}.\left(3^3\right)^5}=\frac{2^{14}.3^{14}}{2^{15}.3^{15}}=\frac{1.1}{2.3}=\frac{1}{6}\)
h) \(\frac{2^{18}.9^4}{6^6.8^4}=\frac{2^{18}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^4}=\frac{2^{18}.3^8}{2^6.3^6.2^{12}}=\frac{2^{18}.3^8}{2^{18}.3^6}=\frac{1.3^2}{1.1}=9\)
o) \(\frac{3^3+3.6^2+6^3}{13}=\frac{3^3+6^2\left(3+6\right)}{13}=\frac{3^3+6^2.3^2}{13}\)
\(=\frac{3^2\left(3+6^2\right)}{13}=\frac{9.3.13}{13}=\frac{9.3.1}{1}=27\)
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
Bài giải
a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)
c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)
d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)
(-10 phần 3 )3.(-6 phần 5)4
=-1000 phần 27.-1296 phần 625
=1296000 phần 16875 bạn tự giuts gọn nhé
2:(1 phần 2 - 2 phần 3)3
=2:(3 phần 6 - 4 phần 6 )3
=2:(-1 phần 6)3
2:-1 phần 216
=2.-216
=-432
63+3.62+33 phần -13
=216+3 . 36+27 phần -13
216+108+27 phần -13
=351 phần -13
=-27
=
Mình rút gọn như sau:
\(\frac{6^3+3.6^2+3^3}{-13}\)
\(=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}\)
\(=\frac{3^2.\left(2^3+3.2^2+3\right)}{-13}\)
\(=\frac{9.\left(8+12+3\right)}{-13}\)
\(=\frac{9.23}{-13}=\frac{207}{-13}\)
Vì tử số và mẫu số ko cùng chia hết cho 1 số nào ngoài 1 suy ra phân số trên tối giản
a.
\(\frac{2^7\times9^3}{6^5\times8^2}=\frac{2^7\times\left(3^2\right)^3}{\left(2\times3\right)^5\times\left(2^3\right)^2}=\frac{2^7\times3^6}{2^5\times3^5\times2^6}=\frac{3}{2^4}=\frac{3}{16}\)
b.
\(\frac{6^3+3\times6^2+3^3}{-13}=\frac{\left(2\times3\right)^3+3\times\left(3\times2\right)^2+3^3}{-13}=\frac{2^3\times3^3+3\times3^2\times2^2+3^3}{-13}=\frac{8\times3^3+3^3\times4+3^3}{-13}\)\(=\frac{3^3\times\left(8+4+1\right)}{-13}=\frac{27\times13}{-13}=-27\)
c.
\(\frac{5^4\times20^4}{25^5\times4^5}=\frac{\left(5\times20\right)^4}{\left(25\times4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
d.
\(\left(\frac{5^4-5^3}{125^4}\right)=\frac{5^3\times\left(5-1\right)}{\left(5^3\right)^4}=\frac{5^3\times4}{5^{12}}=\frac{4}{5^9}\)
a)\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{2^5.3^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}\)
b)\(\frac{6^3+3.6^2+3^3}{-13}=\frac{6.6^2+3.6^2+3^3}{-13}=\frac{6^2.\left(6+3\right)+3^3}{-13}=\frac{6^2.9+3^3}{-13}=\frac{6^2.3^2+3.3^2}{-13}=\frac{3^2.\left(6^2+3\right)}{-13}=\frac{3^2.39}{-13}=3^2.\left(-3\right)=-27\)
c)\(\frac{5^4.20^4}{25^5.4^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
a. \(\frac{20^5.5^{10}}{100^5}\)= \(\frac{20^5.5^{10}}{20^5.5^5}\)= \(5^5\)=\(3125\)
b. \(\frac{0,9^5}{0,3^6}\)= \(\frac{0,9^5}{0,3^5.0,3}\)= \(\left(\frac{0,9}{0,3}\right).\frac{1}{0,3}\)= \(243.\frac{1}{0,3}\)= \(810\)
c.\(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(3.2\right)^3+3.\left(3.2\right)^{^2}+3^3}{-13}=\frac{3^3.2^3+3.3^2.2^2+3^3}{-13}\)\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=3^3.\left(-1\right)=-27\)
\(a,\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.\left(3^2\right)^4}{\left(2.3\right)^6.\left(2^3\right)^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^8}{2^{15}.3^6}=3^2=9\)
\(b,\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}=\frac{3^3.2^3+3.3^2.2^2+3^3}{-13}\)
\(=\frac{3^3.2^3+3^3.2^2+3^3}{-13}=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.\left(8+4+1\right)}{-13}=\frac{3^3.13}{-13}=-3^8=\frac{1}{3^8}\)
\(\frac{6^3+3.6^2+3^3}{-13}=\frac{216+108+27}{-13}=\frac{351}{-13}=-27\)
E = \(\frac{\left(2^2\right)^6.\left(3^2\right) ^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
E = \(\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
E = \(\frac{2^{12}.3^{10}+2^{13}.3^{10}.5}{-2^{11}.3^{11}.\left(2.3+1\right)}\)
E = \(\frac{2^{12}.3^{10}.\left(1+5\right)}{-2^{11}.3^{11}.7}\)
E = \(\frac{2^{12}.3^{10}.6}{-2^{11}.3^{11}.7}\)
E=\(\frac{-2^{11}.\left(-2\right).3^{10}.6}{-2^{11}.3^{10}.3.7}\)
E = \(\frac{-2.6}{3.7}=-\frac{4}{7}\)
Vậy E = -4/7
Ý F bn lm tương tự nha
n=10/7
o=27