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Mình làm nhầm chỗ 29.39.23.3.5 phải bằng 212.310.5. Thông cảm nha bạn
\(\frac{4^6.9^5-6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\frac{2.6}{3.5}\)
\(=\frac{4}{5}\)
a: \(A=\dfrac{5^2\cdot3^{11}\cdot2^{11}\cdot2^8+3^2\cdot2^2\cdot2^{12}\cdot3^6\cdot3^2\cdot5^2}{2\cdot2^{12}\cdot3^{12}\cdot2^4\cdot5^4-3^8\cdot960^3}\)
\(=\dfrac{5^2\cdot3^{11}\cdot2^{19}+3^{10}\cdot2^{14}\cdot5^2}{2^{17}\cdot3^{12}\cdot5^4-3^{11}\cdot2^{18}\cdot5^3}\)
\(=\dfrac{5^2\cdot2^{14}\cdot3^{10}\left(3\cdot2^5+1\right)}{2^{17}\cdot3^{11}\cdot5^3\left(3\cdot5-2\right)}=\dfrac{1}{5}\cdot\dfrac{1}{8}\cdot\dfrac{1}{10}\cdot\dfrac{97}{13}=\dfrac{97}{5200}\)
b: \(B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{12}{15}=\dfrac{4}{5}\)
Giải:
\(10.\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=10.\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=10.\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=10.\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=10.\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\)
\(=10.\dfrac{2^{13}.3^{11}}{2^{11}.3^{11}.5}\)
\(=10.\dfrac{2^2}{5}\)
\(=2^3=8\)
Vậy ...
\(A=\frac{\left(2^2\right)^6\cdot\left(3^2\right)^5+6^9\cdot120}{\left(2^3\right)^4\cdot3^{12}-6^{11}}\)
\(A=\frac{2^{12}\cdot3^{10}+6^9\cdot120}{2^{12}\cdot3^{12}-6^{11}}\)
\(A=\frac{1\cdot1+1\cdot120}{1\cdot3^2-6^2}\)(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
\(10\cdot\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=10\cdot\frac{2^{12}\cdot3^{10}+2\cdot9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=10\cdot\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)
\(=10\cdot\frac{2\cdot6}{3\cdot5}=8\)
Bấm máy là ra nhé bn
812/1215
>< có j k cko mik` vs nkia
\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{6^{12}-6^{11}}\)
\(=\frac{2^{12}3^{10}\left(1+5\right)}{6^{11}\left(6-1\right)}\)
\(=\frac{2^{10}\cdot3^{10}\cdot5\cdot2^2}{6^{10}\cdot6\cdot5}\)
\(=\frac{6^{10}\cdot20}{6^{10}\cdot30}\)
\(=\frac{2}{3}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\) (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)
=\(\frac{2.1.6}{1.3.5}=\frac{2.1.2}{1.1.5}=\frac{4}{5}\)
\(A=\dfrac{4^6.9^5.6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{15}.2^9.3^9.2^3.5.3}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{24}.3^{25}.5}{2^{11}.3^{11}\left(2.3-1\right)}=\dfrac{2^{13}.3^{14}.5}{5}=2^{13}.3^{14}\)
Vậy.......
Chúc bạn học tốt!!!