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\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)= \(\frac{4096.59049+10077696.120}{4096.531441-362797056}\)= \(\frac{241864704+1209323520}{2176782336-362797056}\)= \(\frac{1451188224}{1813985280}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{241864704+1209323520}{2176782336-362797056}=\frac{4}{5}\)
\(A=\)\(\frac{4^6.9^5+6^9.120}{6^{11}+8^4.3^{12}}\)
\(A=\frac{2^{12}.3^{10}+6^9.6.2^2.5}{6^{11}+2^{12}.3^{12}}\)
\(A=\frac{2^{12}.3^{10}+6^{10}.2^2.5}{6^{11}+2^{12}.3^{12}}\)
\(A=\frac{2^{12}.3^{10}+2^{10}.3^{10}.2^2.5}{\left(2.3\right)^{11}+2^{12}.3^{12}}\)
\(A=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}+2^{12}.3^{12}}\)
\(A=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(1+2.3\right)}\)
\(A=\frac{2^{12}.3^{10}.6}{2^{11}.3^{11}.7}\)
\(A=\frac{2^{12}.3^{10}.2.3}{2^{11}.3^{11}.7}\)
\(A=\frac{2^{13}.3^{11}}{2^{11}.3^{11}.7}\)
\(A=\frac{2^2}{7}\)\(=\frac{4}{7}\)
\(\dfrac{4^0\cdot9^3-6^9\cdot1200}{8^4\cdot3^{12}+6^{11}}\)
= \(\dfrac{1\cdot\left(3^2\right)^3-\left(3\cdot2\right)^9\cdot\left(5^2\cdot2^4\cdot3\right)}{\left(2^3\right)^4\cdot3^{12}+6^{11}}\)
= \(\dfrac{3^6-3^9\cdot2^9\cdot5^2\cdot2^4\cdot3}{2^{12}\cdot3^{12}+\left(2\cdot3\right)^{11}}\)
= \(\dfrac{3^6-3^{10}\cdot2^{13}\cdot5^2}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
như lời đã nói thì bn làm tiếp>:)))
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
\(N=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^{10}.4.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}-6^{11}}=\frac{6.2^{12}.3^{10}}{6^{12}-6^{11}}\)
\(=\frac{2.3.2^{12}.3^{10}}{6.6^{11}-6^{11}}=\frac{2^{13}.3^{11}}{5.\left(2.3\right)^{11}}=\frac{2^{13}.3^{11}}{5.2^{11}.3^{11}}=\frac{4}{5}\)
chịu
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{4}{7}\)