Rút gọn: \(A=\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}\)
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\(\frac{9^{14}\cdot25^5\cdot8^7}{18^{12}\cdot625^3\cdot24^3}=\frac{\left(3^2\right)^{14}\cdot\left(5^2\right)^5\cdot\left(2^3\right)^7}{\left(3^2\cdot2\right)^{12}\cdot\left(5^4\right)^3\cdot\left(3\cdot2^3\right)^3}\)
\(=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{3^{24}\cdot2^{12}\cdot5^{12}\cdot3^3\cdot2^9}=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{3^{25}\cdot5^{12}\cdot2^{21}}=\frac{3^3}{5^2}=\frac{27}{25}\)
\(\frac{9^{14}.225^5.8^7}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.\left(3^2.5^2\right)^5.\left(2^3\right)^7}{\left(3^2.2\right)^{12}.\left(5^4\right)^3.\left(3.2^3\right)^3}=\frac{3^{28}.3^{10}.5^{10}.2^{21}}{3^{24}.2^{12}.5^{12}.3^3.2^9}=\frac{3^{38}.5^{10}.2^{21}}{3^{27}.2^{21}.5^{12}}=\frac{3^{11}}{5^2}\)
20112-(304+2012)+(2013+304)
=20112-304-2012+2013+304
=20112+(-2012+2013)+(-304+304)
=20112+1+0=20113
\(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.25^5.\left(2^3\right)^7}{2^{12}.\left(3^2\right)^{12}.\left(25^2\right)^3.\left(2^3\right)^3.3^3}=\)\(\frac{3^{28}.25^5.2^{21}}{2^{12}.2^9.3^{24}.3^3.25^6}=\frac{3^{28}.25^5.2^{21}}{2^{21}.3^{27}.25^6}\)\(=\frac{3}{25}\)
a)\(\frac{\text{3600−75}}{\text{8400−105}}=\frac{3.1200-3.25}{7.1200-7.25}=\frac{3.\left(1200-25\right)}{7.\left(1200-25\right)}=\frac{3}{7}\)
b) \(\frac{9^{14}.25^5.8^2}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.\left(5^2\right)^5.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(2^3.3\right)^3}=\frac{3^{28}.5^{10}.2^{21}}{2^{12}.3^{24}.5^{12}.2^9.3^3}=\frac{3^{28}.5^{10}.2^{21}}{2^{21}.3^{27}.5^{12}}=\frac{3}{5^2}=\frac{3}{25}\)
c) \(C=\frac{1+3+5+...+19}{21+23+25+...+39}=\frac{\left(1+19\right).19:2}{\left(21+39\right).19:2}=\frac{190}{570}=\frac{1}{3}\)
a)3600-75/8400-175
= \frac{75(48 - 1)}{175(48-1)\frac{75(48 - 1)}{175(48-1) = 3737
b)9^14*25^5*8^7/18^12*625^3*24^3
= 328.510.221221.324.512.33.29=352=325328.510.221221.324.512.33.29=352=325
bài 2: tìm n thuộc z để:
a)Phân số A=n+1/n-3 là phân số tối giản
Muốn cho n+1n−3n+1n−3 là phân số tối giản thì (n+1,n-3) = 1 \Rightarrow (n-3,4) =1 \Rightarrow n - 3 không chia hết cho 2 hay n là số chẵn.
bài 3:
tìm x,y thuộc Z sao cho 1/8<x/12<y/9<1/4
Quy đồng mẫu ta được 236236 < 3x363x36< 4y364y36 < 936936
\Rightarrow 2<3x<4y<9
do đó x = 1;y= 1 hoặc x =1;y=2 hoặc x=2;y=2
bài 4:cho phân số A=1+3+5+...+19/21+23+25+...+39
a)rút gọn A
A = \frac{(19+1). 10 : 2}{(39 + 21).10 : 2\frac{(19+1). 10 : 2}{(39 + 21).10 : 2 = 100300=13100300=13
b)hãy xóa 1 số hạng ở tử và 1 số hạng ở mẫu để được một phân số mới vẫn có giá trị bằng A
Gọi số phải xóa ở tử là a; số phải xóa ở mẫu là b
\Rightarrow ab=13ab=13
Vậy ta tìm được (a,b) = {(7,21);(9,27);(11,33};(13,39)}
Ta có: \(\frac{9^9.225^5.8^7}{18^{12}.625^3.24^3}\)=\(\frac{\left(3^2\right)^9.\left(25.3^2\right)^5.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(25^2\right)^3.\left(3.8\right)^3}\)
=\(\frac{3^{18}.25^5.3^{10}.2^{21}}{2^{12}.3^{24}.25^6.3^3.2^9}\)
=\(\frac{3^{28}.25^5.2^{21}}{3^{27}.25^6.2^{21}}\)=\(\frac{3}{25.}\)
\(\dfrac{9^{14}.25^6.8^7}{18^{12}.625^3.24^3}=\dfrac{\left(3^2\right)^{14}.\left(5^2\right)^6.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(2^3.3\right)^3}\)
\(=\dfrac{3^{28}.5^{12}.2^{21}}{2^{21}.3^{27}.5^{12}}=\dfrac{3}{1}=3\)
\(A=\dfrac{9^{14}.25^6.8^7}{18^{12}.625^3.24^3}=\dfrac{\left(3^2\right)^{14}.\left(5^2\right)^6.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(3.2^3\right)^3}\)
=\(\dfrac{3^{28}.5^{12}.2^{21}}{2^{12}.3^{24}.5^{12}.3^3.2^9}\)=\(\dfrac{3^{28}.5^{12}.2^{21}}{2^{21}.3^{27}.5^{12}}=3\)
\(\frac{9^{14}}{18^{12}}.\frac{25^5}{625^3}.\frac{8^7}{24^3}\)
\(=\frac{9^{14}}{\left(9.2\right)^{12}}.\frac{25^5}{25^6}.\frac{8^7}{\left(8.3\right)^3}\)
\(=\frac{9^{14}}{9^{12}.2^{12}}.\frac{1}{25}.\frac{8^7}{8^3.3^3}\)
\(=\frac{9^2}{2^{12}}.\frac{1}{25}.\frac{8^4}{3^3}\)
\(=\frac{81}{4096}.\frac{1}{25}.\frac{4096}{27}\)
\(=\frac{81}{4096}.\frac{4096}{27}.\frac{1}{24}=3.\frac{1}{24}=\frac{3}{24}\)
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