\(A=\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)
\(\dfrac{-4^6.9^5-6^9.120}{-8^4.3^{12}-6^{11}}\)
\(=\dfrac{-2^{12}\cdot3^{10}-2^{12}\cdot3^{10}\cdot5}{-2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2\cdot6}{3\cdot7}=\dfrac{12}{21}=\dfrac{4}{7}\)
Rút gọn biểu thức: \(A=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}=\dfrac{2.6}{3.5}=\dfrac{4}{5}\)
\(A=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+6^9.6.2^2.5}{\left(2^3\right)^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\dfrac{2^{11}.3^{10}\left(2^1+2^1.5\right)}{2^{11}.3^{10}\left(2^1.3^2-1.3^1\right)}=\dfrac{2+10}{2.9-1.3}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot2^3\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot5}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2\cdot6}{3\cdot5}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)
ta có : \(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\dfrac{4\left(4^5.9^5+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}=\dfrac{4\left(2^{10}.3^{10}+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}\)
\(=\dfrac{4\left(6^{10}+5.6^{10}\right)}{-6^{12}-6^{11}}=\dfrac{4.6^{11}}{-6^{11}\left(6+1\right)}=-\dfrac{4}{7}\)
\(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}+6^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{-2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=-\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}=\dfrac{-2}{3}\)
Tính A=\(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)
Bài lớp 7 à bạn khó quá
Mình giải rồi nhưng không ra
treen tử :phân tích ra thừa số nguyên tố rồi đặt thừa số chung vs số mũ nhỏ hơn
dưới mẫu : âm nhân âm = dương , phân tích ra thừa số nguyên tố
Sau đó rut gọn ra kết quả
Câu 2: Tính
C = \(\dfrac{6^3+3.6^2+3^3}{13}\)
D = \(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(C=\dfrac{6^3+3\cdot6^2+3^3}{13}=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}=27\)
.\(10.\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
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 ...
\(\dfrac{10^3+2.5^3+5^3}{55}-\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{5^3\left(2^3+2+1\right)}{55}-\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=5^2-\dfrac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=25-\dfrac{2}{3}\cdot\dfrac{6}{5}\)
=25-4/5
=24,2
Tính
a, \(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)
\(=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{2^{12}.3^{10}+3^9.2^9.3.2^3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}3^{11}\left(6-1\right)}\)
\(=\dfrac{2^{13}3^{11}}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^2}{5}=\dfrac{4}{5}\)