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\(\dfrac{9^{14}.25^6.8^7}{18^{12}.625^3.24^3}=\dfrac{\left(3^2\right)^{14}.\left(5^2\right)^6.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(2^3.3\right)^3}\)
\(=\dfrac{3^{28}.5^{12}.2^{21}}{2^{21}.3^{27}.5^{12}}=\dfrac{3}{1}=3\)
\(A=\dfrac{9^{14}.25^6.8^7}{18^{12}.625^3.24^3}=\dfrac{\left(3^2\right)^{14}.\left(5^2\right)^6.\left(2^3\right)^7}{\left(2.3^2\right)^{12}.\left(5^4\right)^3.\left(3.2^3\right)^3}\)
=\(\dfrac{3^{28}.5^{12}.2^{21}}{2^{12}.3^{24}.5^{12}.3^3.2^9}\)=\(\dfrac{3^{28}.5^{12}.2^{21}}{2^{21}.3^{27}.5^{12}}=3\)
\(\frac{9^{14}\cdot25^5\cdot8^7}{18^{12}\cdot625^3\cdot24^3}=\frac{\left(3^2\right)^{14}\cdot\left(5^2\right)^5\cdot\left(2^3\right)^7}{\left(3^2\cdot2\right)^{12}\cdot\left(5^4\right)^3\cdot\left(3\cdot2^3\right)^3}\)
\(=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{3^{24}\cdot2^{12}\cdot5^{12}\cdot3^3\cdot2^9}=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{3^{25}\cdot5^{12}\cdot2^{21}}=\frac{3^3}{5^2}=\frac{27}{25}\)
\(\frac{9^{14}}{18^{12}}.\frac{25^5}{625^3}.\frac{8^7}{24^3}\)
\(=\frac{9^{14}}{\left(9.2\right)^{12}}.\frac{25^5}{25^6}.\frac{8^7}{\left(8.3\right)^3}\)
\(=\frac{9^{14}}{9^{12}.2^{12}}.\frac{1}{25}.\frac{8^7}{8^3.3^3}\)
\(=\frac{9^2}{2^{12}}.\frac{1}{25}.\frac{8^4}{3^3}\)
\(=\frac{81}{4096}.\frac{1}{25}.\frac{4096}{27}\)
\(=\frac{81}{4096}.\frac{4096}{27}.\frac{1}{24}=3.\frac{1}{24}=\frac{3}{24}\)
**** **** ****
20112-(304+2012)+(2013+304)
=20112-304-2012+2013+304
=20112+(-2012+2013)+(-304+304)
=20112+1+0=20113
\(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{\left(3^2\right)^{14}.25^5.\left(2^3\right)^7}{2^{12}.\left(3^2\right)^{12}.\left(25^2\right)^3.\left(2^3\right)^3.3^3}=\)\(\frac{3^{28}.25^5.2^{21}}{2^{12}.2^9.3^{24}.3^3.25^6}=\frac{3^{28}.25^5.2^{21}}{2^{21}.3^{27}.25^6}\)\(=\frac{3}{25}\)
đặt 2n + 34 = a^2
34 = a^2-n^2
34=(a-n)(a+n)
a-n thuộc ước của 34 là { 1; 2; 17; 34} và a-n . Ta có bảng sau ( mik ko bt vẽ)
=> a-n 1 2
a+n 34 17
Mà tổng và hiệu 2 số nguyên cùng tính chẵn lẻ
Vậy ....
Ta cóS = 14 +24 +34 +···+1004 không là số chính phương.
=> S= (1004+14).100:2=50 900 ko là SCP
a)\(2^{2n-1}+4^{n+2}=264\)
\(264=2^3\cdot3\cdot11\)
\(2^3=2^{\left(3+1\right)\div2}=2^2\Rightarrow n=2\)
\(4^{n+2}=264-2^3=256\)
\(256=4^4=4^{4-2}=4^2\Rightarrow n=2\)
vậy \(n=2\)
b) \(P=\frac{9^{14}\cdot25^6\cdot8^7}{18^{12}\cdot625^3\cdot24^3}\)
\(P=\frac{9^{14}\cdot25^6\cdot8^7}{18^{12}\cdot25^6\cdot25^6\cdot24^3}\)
\(P=\frac{9^{14}\cdot8^7}{18^{12}\cdot24^3}=3\)