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\(10\cdot\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=10\cdot\frac{2^{12}\cdot3^{10}+2\cdot9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=10\cdot\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)
\(=10\cdot\frac{2\cdot6}{3\cdot5}=8\)
\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{6^{12}-6^{11}}\)
\(=\frac{2^{12}3^{10}\left(1+5\right)}{6^{11}\left(6-1\right)}\)
\(=\frac{2^{10}\cdot3^{10}\cdot5\cdot2^2}{6^{10}\cdot6\cdot5}\)
\(=\frac{6^{10}\cdot20}{6^{10}\cdot30}\)
\(=\frac{2}{3}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\) (Sau đó phân tách ra)
=\(\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
=\(\frac{2^{12}.3^{10}+2^9.3^9.2^33.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
=\(\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\) (Gộp và giải như biểu thức thường)
=\(\frac{2^{12}.3^{10}.\left(1+1.5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
=\(\frac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\) (Rút gọn giữa tử và mẫu)
=\(\frac{2.1.6}{1.3.5}=\frac{2.1.2}{1.1.5}=\frac{4}{5}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}.10=\frac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}-2^{11}.3^{11}}.10=\frac{2^{10}.3^{10}\left(2^2+20\right)}{2^{10}.3^{10}\left(6^2-6\right)}.10=\frac{24.10}{30}=8\)
\(A=\frac{\left(2^2\right)^6\cdot\left(3^2\right)^5+6^9\cdot120}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\)
\(A=\frac{2^{12}\cdot3^{10}+6^9\cdot120}{2^{12}\cdot3^{12}-6^{11}}\)
\(A=\frac{1\cdot1+1\cdot120}{1\cdot3^2-6^2}\)(Ở BƯỚC NÀY MÌNH RÚT GỌN TỰ VỚI MẪU VÌ HAI SỐ CÙNG CƠ SỐ)
\(A=\frac{1+120}{1\cdot9-36}\)
\(A=\frac{121}{9-36}\)
\(A=\frac{121}{-27}=-\frac{121}{27}\)
NHỚ NHA