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A=1.5.(3.2)+2.10.(6.2)+3.15.(9.2)+4.20.(12.2)+5.25.(15.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=1.5.3+2.10.6+3.15.9+4.20.12+5.25.15(2.2.2.2.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=2.2.2.2.2
A=32
\(\frac{1\cdot3\cdot5\cdot2+2\cdot10\cdot6\cdot2+3\cdot15\cdot9\cdot2+4\cdot20\cdot12\cdot2+5\cdot25\cdot15\cdot2}{1\cdot3\cdot5+2\cdot10\cdot6+3\cdot15\cdot9+4\cdot20\cdot12+5\cdot25\cdot15 }\)
\(2\cdot2\cdot2\cdot2\cdot2=2^5\)
\(=32\)
\(a=\frac{6.2.3.4+6.3.4.5+6.4.5.6+...+6.98.99.100}{2.3.4+3.4.5+4.5.6+...+98.99.100}=6\)
thay vào p(x) suy ra a không là nghiệm của đa thức nhé bạn
\(a,\frac{20^{12}\cdot6^{14}}{8^{13}\cdot15^{12}}\)
\(=\frac{5^{12}\cdot2^{24}\cdot2^{14}\cdot3^{14}}{2^{39}\cdot3^{12}\cdot5^{12}}\)
\(=\frac{5^{12}\cdot2^{38}\cdot3^{14}}{2^{39}\cdot3^{12}\cdot5^{12}}=\frac{3^2}{2}=\frac{9}{2}\)
\(b,\frac{45^{12}\cdot10^{14}}{18^{13}\cdot25^{12}}\)
\(=\frac{5^{12}\cdot3^{24}\cdot2^{14}\cdot5^{14}}{2^{13}\cdot3^{26}\cdot5^{24}}\)
\(=\frac{5^{26}\cdot3^{24}\cdot2^{14}}{2^{13}\cdot3^{26}\cdot5^{24}}=\frac{5^2\cdot2}{3^2}=\frac{50}{9}\)
\(c,\frac{18^{12}\cdot27^8}{6^8\cdot3^{40}}\)
\(=\frac{2^{12}\cdot3^{24}\cdot3^{24}}{2^8\cdot3^8\cdot3^{40}}\)
\(=\frac{2^{12}\cdot3^{48}}{2^8\cdot3^{48}}=2^4=16\)
\(d,\frac{12^{14}\cdot9^{18}}{8^9\cdot27^{17}}\)
\(=\frac{3^{14}\cdot2^{28}\cdot3^{36}}{2^{27}\cdot3^{51}}\)
\(=\frac{3^{50}\cdot2^{28}}{2^{27}\cdot3^{51}}=\frac{2}{3}\)
làm hơi tắt nên chịu khó hiểu
\(\frac{7.2^{32}.3^8.5^4-2^9.3^9.2^8.5^4}{2^{10}.3^{10}.2^5.5^5-7.3^9.4^8.5^4}=\frac{7.2^{32}.3^8.5^4-2^{17}.3^9.5^4}{2^{15}.3^{10}.5^5-7.3^9.2^{16}.5^4}\)
\(\frac{2^{17}.3^8.5^4\left(2^5.7-1\right)}{2^{15}.3^9.5^4\left(3.5-7.2\right)}=\frac{2^{17}.3^8.5^4\left(32.7-1\right)}{2^{15}.3^9.5^4\left(15-14\right)}\)
\(S=\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)-\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}\right)\)
=>\(S=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\right)-\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}\right)\)
=>\(S=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}\right)\)
=>\(S=\frac{1}{2}.\left(1-\frac{1}{9}\right)-\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{10}\right)\)
=>\(S=\frac{1}{2}.\frac{8}{9}-\frac{1}{2}.\frac{2}{5}\)
=>\(S=\frac{4}{9}-\frac{1}{5}\)
=>\(S=\frac{11}{45}\)
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=-\frac{1}{3}\)
Ta có:\
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(A=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(A=-\frac{2}{6}=-\frac{1}{3}\)
\(M=\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(M=\frac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(M=\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(M=\frac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(M=\frac{-2}{6}=\frac{-1}{3}\)
=\(\frac{30\left(1+8+27+64+125\right)}{5\left(3+24+81+64+375\right)}\)
= \(\frac{30.225}{5.574}\)
=\(\frac{6750}{2870}\)
=\(\frac{675}{287}\)
K mình với!!!!