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\(\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}\)
\(=\frac{5^2.2^{11}.3^{11}.2^8+2^2.3^2.2^6.2^6.3^6.3^2.5^2}{2.2^{12}.3^{12}.2^4.5^4-3^4.2^{18}.3^3.5^3}\)
\(=\frac{5^2.2^{19}.3^{11}+2^{14}.3^{10}.5^2}{2^{17}.3^{12}.5^4-3^7.2^{18}.5^3}\)
\(=\frac{5^2.2^{14}.3^{10}.\left(2^5.3-1\right)}{2^{17}.3^7.5^3.\left(3^5.5-2\right)}\)
\(=\frac{3^3.95}{2^3.5.1213}\)
\(=\frac{2565}{48520}\)\(=\frac{513}{9704}\)
Bạn Aoi Ogata trả lời sai rồi mình bấm máy tính kết quả khác cơ nhưng mình vẩn cho bạn một k nhé
Bài 1:
a) 6/2 x+ 1 = 2/7
6/2 x = 2/7 - 1
6/2 x = 2/7 - 7/7
6/2 x = -5/7
x = - 5/7 : 6/2
x = - 5/7 . 2/6
x = -5/21
Ủng hộ nha! :)
BÀI 1 a 6/2x+1=2/7
6/2x=2/7-1
6/2x=-5/7
6*7=5*2x
42=5*2x
42/5=2x
x=42/5:2
x=21/5
bai 3
\(A=\frac{10^{2004}+1}{10^{2005}+1}\)
\(10A=\frac{10^{2004}+10}{10^{2005}+1}\)
\(10A=1\frac{9}{10^{2005}+1}\)
\(B=\frac{10^{2005}+1}{10^{2006}+1}\)
\(10B=\frac{10^{2005}+10}{10^{2006}+1}\)
\(10B=1\frac{9}{10^{2006}+1}\)
Vì \(1\frac{9}{10^{2005}+1}>1\frac{9}{10^{2006}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
bai 4
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^8}\)
\(\frac{1}{3}A=\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+....+\frac{1}{3^9}\)
\(A-\frac{1}{3}A=\frac{1}{3}-\frac{1}{3^9}\)
\(\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}\)
\(=\frac{2^{19}.5^2.3^{11}+2^{14}.3^{10}.5^2}{2^{17}.3^{12}.5^4-2^{18}.3^{11}.5^3}\)
\(=\frac{2^{14}.3^{10}.5^2\left(2^5.3+1\right)}{2^{17}.3^{11}.5^3\left(3.5-2\right)}=\frac{97}{2^3.3.5.13}=\frac{97}{1560}\)