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\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^4.7^4}{5^9.7^3+5^92^3.7^3}\)\(=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\frac{7^3\left(5^{10}-5^4.7^4\right)}{7^3\left(5^9+5^9.2^3\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}-\frac{5^{10}-5^4.7^4}{5^9+5^9.2^3}\)\(=\frac{3^4\left(3-1\right)}{3^4\left(3^2+3\right)}-\frac{5^4\left(5^6-7^4\right)}{5^4\left(5^5+5^5.2^3\right)}\)
\(=\frac{1}{6}-\frac{5^6-7^4}{5^5+5^5.2^3}\)cậu cứ phân tích từ từ
không vt lại đề
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^6.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}\cdot3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^6.7^3\left(1+2^3\right)}\)
\(=\frac{1}{6}-\frac{-10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\frac{5^{10}.7^4-25^5.49^2}{\left(125.7\right)3+5^9.\left(14\right)^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{125^3.7^3+5^9.\left(2.7\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+8\right)}\)
\(=\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}=\frac{2}{12}-\frac{-30}{9}\)
\(=\frac{1}{6}+\frac{10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}=\frac{7}{2}\)
Bạn ơi cho mk hỏi chỗ đoạn kia bạn lấy 1-7 ở đâu và 1 + 8 ở đâu
a=2^12.3^5-2^12.3^4/2^12.3^6+2^12.3^5 - 5^10.7^3-5^10.7^4/5^9.7^3+5^9.7^3.2^3
a=2^12.3^4.(3-1)/2^12.3^5.(3+1)-5^10.7^3.(1-7)/5^9.7^3.(1+8)
a=2/12-30/9
a=1/6-10/3=-19/6
a=
\(=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{18}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)
\(=\dfrac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\left(2^6\cdot3+1\right)}-\dfrac{5^{10}\cdot7^3\cdot\left(-6\right)}{5^9\cdot7^3\cdot9}\)
\(=\dfrac{2}{193}+\dfrac{5\cdot2}{3}=\dfrac{1936}{579}\)
1/ \(\frac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)
\(=\frac{5^{20}.3^{29}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}=\frac{3^{29}.5^{18}.\left(25-9\right)}{3^{29}.5^{18}.\left(7-5\right)}=\frac{16}{2}=8\)
CÁC BÀI CÒN LẠI TƯƠNG TỰ HẾT NHÉ E
mk ko viết lại đề
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}+\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}+\frac{2^{12}.3^{10}\left(1+5\right)}{2.\left(2^{12}.3^{12}\right)}\)
\(=\frac{2}{3.4}+\frac{2^{12}.3^{10}.6}{2.2^{12}.3^{12}}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)
Vậy A= \(\frac{1}{2}\)