a) \(\dfrac{1}{9}.27^n=3^n\)

\(\Leftrightarrow\dfrac{1}{9}=3^n:27^n\)

\(\Leftrightarrow\dfrac{1}{9}=\left(\dfrac{3}{27}\right)^n\)

\(\Leftrightarrow\dfrac{1}{9}=\left(\dfrac{1}{9}\right)^n\)

\(\Leftrightarrow n=1\)

b) \(3^{-2}.3^4.3^n=3^7\)

\(\Leftrightarrow3^2.3^n=3^7\)

\(\Leftrightarrow3^n=3^7:3^2\)

\(\Leftrightarrow3^n=3^5\)

\(\Leftrightarrow n=5\)

c) \(32^{-n}.16^n=2048\)

\(\Leftrightarrow\left(2^5\right)^{-n}.\left(2^4\right)^n=2^{11}\)

\(\Leftrightarrow2^{-5n}.2^{4n}=2^{11}\)

\(\Leftrightarrow2^{-n}=2^{11}\)

\(\Leftrightarrow n=-11\)

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

\(N=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^{10}.4.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

    \(=\frac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}-6^{11}}=\frac{6.2^{12}.3^{10}}{6^{12}-6^{11}}\)

    \(=\frac{2.3.2^{12}.3^{10}}{6.6^{11}-6^{11}}=\frac{2^{13}.3^{11}}{5.\left(2.3\right)^{11}}=\frac{2^{13}.3^{11}}{5.2^{11}.3^{11}}=\frac{4}{5}\)

\(a=lim\frac{\left(\frac{2}{3}\right)^n+1}{3\left(\frac{1}{3}\right)^n-12}=-\frac{1}{12}\)

\(b=lim\frac{4\left(\frac{4}{10}\right)^n+1}{\left(\frac{3}{10}\right)^n-40}=-\frac{1}{40}\)

\(c=lim\frac{1-\left(\frac{2}{12}\right)^n}{1+45\left(\frac{3}{12}\right)^n}=\frac{1}{1}=1\)

\(d=\frac{\left(-\frac{2}{3}\right)^n+1}{-2\left(-\frac{2}{3}\right)^n-12+2\left(\frac{1}{3}\right)^n}=-\frac{1}{12}\)

\(e=\frac{1-11\left(\frac{1}{3}\right)^n}{\left(\frac{1}{3}\right)^n+14\left(\frac{2}{3}\right)^n}=\frac{1}{0}=+\infty\)

\(f=\frac{\left(\frac{2}{5}\right)^n-3+\left(\frac{1}{5}\right)^n}{3\left(\frac{2}{5}\right)^n+28\left(\frac{4}{5}\right)^n}=\frac{-3}{0}=-\infty\)

\(A=\frac{15.3^{11}+4.27^1}{9^7}\)

\(\Rightarrow A=\frac{3.5.3^{11}+4.3^{3^1}}{\left(3^2\right)^7}\)

\(\Rightarrow A=\frac{3^{12}.5+4.3^3}{3^{14}}\)

\(\Rightarrow A=\frac{3^3.\left(5.3^8+4.3^3\right)}{3^{14}}\)

\(\Rightarrow A=\frac{32805+4}{177147}\)

\(\Rightarrow A=\frac{32809}{177147}\)

Câu 1 : \(1,321338308x10^{-4}\)

Câu 2 : \(1316,572106\)

Câu 3 : \(1,641302619x10^{-13}\)

Ủng hộ nhé,tớ đang âm.

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