Tính \(\frac{35.\left(27^8+2.9^{11}\right)}{15.\left(81^6-12.3^{19}\right)}\)
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HN
0
CN
1
13 tháng 5 2017
\(\frac{35\left(27^8+2.9^{11}\right)}{15\left(81^6-12.3^{19}\right)}\)= \(\frac{35\left(\left(3^3\right)^8+2.\left(3^2\right)^{11}\right)}{15\left(\left(3^4\right)^6-4.3.3^{19}\right)}\)= \(\frac{35\left(3^{24}+2.3^{22}\right)}{15\left(3^{24}-4.3^{20}\right)}\)= \(\frac{35\left(3^{22}.3^2+2.3^{22}\right)}{15\left(3^{20}.3^4-4.3^{20}\right)}\)= \(\frac{35\left(3^{22}.\left(9+2\right)\right)}{15\left(3^{20}.\left(81-4\right)\right)}\)= \(\frac{35\left(3^{22}.11\right)}{15\left(3^{20}.77\right)}\)= \(\frac{5.7.3^{22}.11}{5.3.3^{20}.7.11}\)= \(\frac{3^{22}}{3.3^{20}}\)= \(\frac{3^{20}.3.3}{3.3^{20}}\)= \(\frac{3}{1}\)= 3
PT
0
26 tháng 11 2017
Bài 1 : chị phân tích ra thừa số nguyên tố, rồi rút gọn đi là ok mak
Bài 2:
\(B=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)........\left(12^4+\dfrac{1}{4}\right)}\)
\(=\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right).........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+1+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right).......\left(12^2-12+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......... \left(12.13+\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)
\(=\dfrac{1}{313}\)
26 tháng 11 2017
\(A=\dfrac{35.\left(27^8+2.9^{11}\right)}{15.\left(81^6-12.3^{19}\right)}\)
\(=\dfrac{35.27^8+35.2.9^{11}}{15.81^6-15.12.3^{19}}\)
\(=\dfrac{5.7.\left(3^3\right)^8+5.7.\left(3^2\right)^{11}}{3.5.\left(3^4\right)^6-3.5.3.2^2.3^{19}}\)
\(=\dfrac{5.7.3^{24}+5.7.3^{22}}{5.3^{25}-3^{21}.2^2.5}\)
\(=\dfrac{5.7.3^{22}\left(3^2+1\right)}{5.3^{21}\left(3^4-2^2\right)}\)
\(=\dfrac{7.2.10}{81-4}\)
\(=\dfrac{720}{77}\)
\(\frac{35.\left(27^8+2.9^{11}\right)}{15\left(81^6-12.3^{19}\right)}=\frac{5.7\left(3^{24}+2\cdot3^{22}\right)}{3.5\left(3^{24}-2^2.3^{20}\right)}\)