\(\frac{4^6\cdot9^5+69\cdot120}{8^4\cdot31^2-6^{11}}\)rút gọn phân số (tối giản luôn nhé)
Tính :\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
(Nhập kết quả dưới dạng phân số tối giản)
\(=\frac{241864704+1209323520}{2176782336-362797056}\)
\(=\frac{1451188224}{1813985280}\)
\(=\frac{4}{5}\)
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\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^2-6^{11}}\)
. là nhân nhé . Đề bài thính hợp lý
nhanh gọn lẹ cho mik nge
\(\frac{4^6.9^5+6^9.120}{8^4.3^2-6^{11}}\)
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^2-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^2.\left(2-3^9\right)}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^2.\left(2-3^9\right)}\)
\(=\frac{2.3^8.6}{2-3^9}\)
\(=\frac{2.2.3.3^8}{2-3^9}\)
\(=\frac{2^2.3^9}{2-3^9}\)
RÚT GỌN
a/\(\frac{9^4\cdot27^5\cdot3^6\cdot3^4}{3^8\cdot81^4\cdot234\cdot8^6}\)
b/\(N=\frac{4^6\cdot9^5+6^6\cdot120}{8^4-3^{12}-6}\)
1.Rút gọn(nếu cần) rồi so sánh
\(\frac{\left(-5\right)^2-5\cdot3^2}{5^3+5^2\cdot3^2}\) ;\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\) và\(\frac{2929-101}{2\cdot1919+404}\)
Bài 1 Cho biểu thức
P= \(\frac{6n+5}{2n-4}\)
a) Với giá trị nào của n thì P là phân số
b) Tìm n thuộc Z để P thuộc Z
c) Tính P khi |2n-3|=\(\frac{5}{3}\)
Bài 2 Rút gọn phân số
a) M= \(\frac{9^4\cdot27^5\cdot3^6\cdot3^4}{3^8\cdot81^4\cdot23^4\cdot8^2}\)
b) N= \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
Ta có :
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.23^4.8^2}\)
\(M=\frac{\left(3^2\right)^4.\left(3^3\right)^5.3^{10}}{3^8.\left(3^4\right)^4.23^4.8^2}\)
\(M=\frac{3^8.3^{15}.3^{10}}{3^8.3^{16}.23^4.8^2}\)
\(M=\frac{3^{33}}{3^{24}.23^4.8^2}\)
\(M=\frac{3^9}{23^4.8^2}\)
Bài 1
a) \(P=\frac{6n+5}{2n-4}=\frac{6n-12+7}{2n-4}=3+\frac{7}{2n-4}\)
Để P là phân số thì \(\hept{\begin{cases}2n-4\ne7\\2n-4\ne1\end{cases}}\Leftrightarrow\hept{\begin{cases}n\ne\frac{11}{2}\\n\ne\frac{5}{2}\end{cases}}\)
Vậy...
b) \(P=\frac{6n+5}{2n-4}=3+\frac{7}{2n-4}\)
Để \(P\in Z\)thì \(\orbr{\begin{cases}2n-4=7\\2n-4=1\end{cases}\Leftrightarrow\orbr{\begin{cases}n=\frac{11}{2}\notin Z\\n=\frac{5}{2}\notin Z\end{cases}}}\)
Vậy không có giá trị n nào thuộc Z để P thuộc Z.
c) \(\left|2n-3\right|=\frac{5}{3}\)
Trường hợp: \(2n-3=\frac{5}{3}\Rightarrow n=\frac{7}{3}\)
\(P=\frac{6.\frac{7}{3}+5}{2.\frac{7}{3}-4}=\frac{19}{\frac{2}{3}}=\frac{57}{2}\)
Trường hợp: \(2n-3=-\frac{5}{3}\Rightarrow n=\frac{2}{3}\)
\(P=\frac{6.\frac{2}{3}+5}{2.\frac{2}{3}-4}=\frac{9}{\frac{-8}{3}}=\frac{27}{-8}\)
Bài 2
\(N=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^{10}.4.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}-6^{11}}=\frac{6.2^{12}.3^{10}}{6^{12}-6^{11}}\)
\(=\frac{2.3.2^{12}.3^{10}}{6.6^{11}-6^{11}}=\frac{2^{13}.3^{11}}{5.\left(2.3\right)^{11}}=\frac{2^{13}.3^{11}}{5.2^{11}.3^{11}}=\frac{4}{5}\)
Tính
\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}}{2^{11}.3^{11}}.\frac{1+5}{2.3-1}\)
\(=\frac{2^{12}.3^{10}}{2^{11}.3^{11}}.\frac{6}{5}\)
\(=\frac{2}{3}.\frac{6}{5}\)
\(=\frac{4}{5}\)
\(=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.8.3.5}{\left(2^3\right)^4.3^{12}+\left(2.3+1\right)}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)} \)
\(\frac{2.3.6}{7}=\frac{36}{7}\)
Mình ghi mỏi tay
\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}\cdot6^{11}}\)
Tính: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{4}{5}\)
Tính A, biết:
\(A=\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(A=\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot2^3\cdot3\cdot5}{\left(2^3\right)^4\cdot3^{12}-\left(2\cdot3\right)^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\frac{2\left(1+5\right)}{3\left(6-1\right)}=\frac{2\cdot6}{3\cdot5}=\frac{2\cdot2}{5}=\frac{4}{5}\)
\(\approx GOOD\)\(LUCK\approx\)