TÍNH
a) \(\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}\)
b)\(\frac{25^8+25^{24}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)
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Câu a) bạn chỉ cần phân tích ra rồi rút gọn
Câu b)Ta có công thức cấp số nhân: S=U1.(qn-1)/q-1
Thay vào daỹ số ta đc:
S=1.((254)8-1)/254-1=2532-1/254-1
Bài 1:
a) \(\frac{5^2.6^{11}.16^2+6^2.11^6.15^2}{2.6^{12}.10^{14}-81^2.960^3}=\frac{5^2.2^{19}.3^{11}+2^2.3^4.11^6.5^2}{2^{27}.3^{12}.5^{14}-3^{11}.2^{18}.5^3}\)
\(=\frac{5^2.2^2.3^4.\left(2^{17}.3^7+11^6\right)}{2^{18}.3^{11}.5^3.\left(2^9.3.5^{11}-1\right)}=\frac{2^{17}.3^7+11^6}{2^{16}.3^7.5.\left(2^9.3.5^{11}-1\right)}\)
b) Đặt A = 2528 + 2524 +....+ 254 +1
=> 254.A = 2532 + 2528 +...+ 258 + 254
=> 254.A - A = 2532 -1
\(A=\frac{25^{32}-1}{25^4-1}\)
tương tự....
Thay vào:
\(\frac{25^{28}+25^{24}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}=\frac{\frac{25^{32}-1}{25^4-1}}{\frac{25^{32}-1}{25^2-1}}=\frac{25^2-1}{25^4-1}\)
Ta có: \(B=\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}=\frac{5^2.\left(2.3\right)^{11}.\left(4^2\right)^2+\left(2.3\right)^2.\left(2^2.3\right)^6.\left(3.5\right)^2}{2.\left(2.3\right)^{12}.\left(2.5\right)^4-\left(3^4\right)^2.\left(2^6.3.5\right)^3}\)
\(\frac{5^2.2^{11}.3^{11}.2^8+2^2.3^2.2^{12}.3^6.3^2.5^2}{2.2^{12}.3^{12}.2^4.5^4-3^8.2^{18}.3^3.5^3}=\frac{5^2.2^{19}.3^{11}+2^{14}.3^{10}.5^2}{2^{17}.3^{12}.5^4-3^{11}.2^{18}.5^3}\)
\(=\frac{5^2.2^{14}.3^{10}.\left(2^5.3+1\right)}{2^{17}.3^{11}.5^3.\left(3.5-2\right)}=\frac{2^{14}.3^{10}.5^2.97}{2^{17}.3^{11}.5^3.13}=\frac{1.1.1.97}{2^3.3.5.13}=\frac{97}{1560}\)
\(B=\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}\)
\(B=\frac{5^2.2^{11}.3^{11}.2^{4^2}+2^2.3^2.2^{2^6}.3^6.3^2.5^2}{2.2^{12}.3^{12}.10^4-3^{4^2}.2^{5^3}.3^3}\)
\(B=\frac{5^2.2^{11}.2^8.3^{11}+2^2.2^{12}.3^2.3^6.3^2.5^2}{2^{13}.3^{12}.10^4-3^8.3^3.2^{15}}\)
\(B=\frac{5^2.2^{19}.3^{11}+2^{14}.3^{10}.5^2}{2^{13}.3^{12}.10^4-3^{11}.2^{15}}\)
\(B=\frac{5^2.2^{14}.3^{10}.\left(2^5.3+1\right)}{2^{13}.3^{11}.\left(3.10^4-2^2\right)}\)
\(B=\frac{5^2.2^{14}.3^{10}.97}{2^{13}.3^{11}.29996}\)
\(B=\frac{5^2.2.97}{3.29996}\)
\(B=\frac{4850}{89988}\)
\(B=\frac{2425}{44994}\)
http://olm.vn/hoi-dap/question/425436.html
\(\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}=\frac{5^2.6^{11}.6^2.6^2.6^6.2^6.3^2.5^2}{2.2^{12}.3^{12}2^4.5^4-3^8.2^{15}.3^3}=\frac{5^6.6^{21}.2^8.3^2}{2^{17}.3^{12}.5^4-3^{11}.2^{15}}=\frac{5^6.3^{23}.2^{29}}{3^{11}.2^{15}.\left(2^2.3.5^4-1\right)}=\frac{5^6.3^{13}.2^{14}}{2^2.3.5^4-1}\)