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27 tháng 11 2017

4) \(3^{n+2}+3^n=270\)

\(\Rightarrow3^n.3^2+3^n=270\)

\(\Rightarrow3^n.\left(3^2+1\right)=270\)

\(\Rightarrow3^n.\left(9+1\right)=270\)

\(\Rightarrow3^n.10=270\)

\(\Rightarrow3^n=270:10\)

\(\Rightarrow3^n=27\)

\(\Rightarrow3^n=3^3\)

\(\Rightarrow n=3\)

Vậy \(n=3\)

2 tháng 12 2017

Ta có: \(A=\dfrac{3^6.21^{12}}{175^9.7^3}=\dfrac{3^6.3^{12}.7^{12}}{5^{18}.7^9.7^3}=\dfrac{3^{18}.7^{12}}{5^{18}.7^{12}}=\dfrac{3^{18}}{5^{18}}=\left(\dfrac{3}{5}\right)^{18}\)

\(B=\dfrac{3^{10}.6^7.4}{10^9.5^8}=\dfrac{3^{10}.2^7.3^7.2^2}{2^9.5^9.5^8}=\dfrac{3^{17}.2^9}{2^9.5^{17}}=\dfrac{3^{17}}{5^{17}}=\left(\dfrac{3}{5}\right)^{17}\)

\(\left(\dfrac{3}{5}\right)^{18}< \left(\dfrac{3}{5}\right)^{17}\Rightarrow A< B\)

Vậy A < B

22 tháng 9 2017

Đặt\(A=\dfrac{3^6.21^{12}}{175^9.7^3}=\dfrac{3^6.3^{12}.7^{12}}{175^9.7^3}=\dfrac{3^{18}.7^{12}}{\left(5^2\right)^9.7^9.7^3}=\dfrac{3^{18}.7^{12}}{5^{18}.7^{12}}=\dfrac{3^{18}}{5^{18}}=\left(\dfrac{3}{5}\right)^{18}\)

Đặt \(B=\dfrac{3^{10}.6^7.4}{10^9.5^8}=\dfrac{3^{10}.3^7.2^7.2^2}{2^9.5^9.5^8}=\dfrac{3^{17}.2^9}{2^9.5^{17}}=\dfrac{3^{17}}{5^{17}}=\left(\dfrac{3}{5}\right)^{17}\)

\(\left(\dfrac{3}{5}\right)^{18}>\left(\dfrac{3}{5}\right)^{17}\Leftrightarrow A>B\)

\(\Rightarrow\dfrac{3^6.21^{12}}{175^9.7^3}>\dfrac{3^{10}.6^7.4}{10^9.5^8}\)

22 tháng 8 2016

Đặt: \(A=\frac{3^6.21^{12}}{175^9.7^3}=\frac{3^{18}.7^{12}}{7^{12}.25^9}=\frac{3^{18}}{5^{18}}=\left(\frac{3}{5}\right)^{18}\)

\(B=\frac{3^{10}.6^7.4}{10^9.5^8}=\frac{3^{10}.2^7.3^7.2^2}{2^9.5^9.5^8}=\frac{3^{17}.2^9}{2^9.5^{17}}=\left(\frac{3}{5}\right)^{17}\)

Vì: \(\left(\frac{3}{5}\right)^{18}< \left(\frac{3}{5}\right)^{17}\Rightarrow A< B\)

22 tháng 8 2016

cảm ơn bạn nhiều nha!

15 tháng 9 2019

Bài 1 :                                                                        Bài giải

\(\frac{28^{15}\cdot3^{17}}{84^{16}}=\frac{\left(2^2\cdot7\right)^{15}\cdot3^{17}}{\left(2^2\cdot3\cdot7\right)^{16}}=\frac{2^{30}\cdot7^{15}\cdot3^{17}}{2^{32}\cdot3^{16}\cdot7^{16}}=\frac{3}{2^2\cdot7}=\frac{3}{4\cdot7}=\frac{3}{28}\)

Bài 2 :                                                              Bài giải

\(\frac{3^6\cdot21^{12}}{175^9\cdot7^3}=\frac{3^6\cdot\left(3\cdot7\right)^{12}}{\left(5^2\cdot7\right)^9\cdot7^3}=\frac{3^6\cdot3^{12}\cdot7^{12}}{5^{18}\cdot7^9\cdot7^3}=\frac{3^{18}\cdot7^{12}}{5^{18}\cdot7^{12}}=\frac{3^{18}}{5^{18}}\)

\(\frac{3^{10}\cdot6^7\cdot4}{10^9\cdot5^8}=\frac{3^{10}\cdot\left(2\cdot3\right)^7\cdot2^2}{\left(2\cdot5\right)^9\cdot5^8}=\frac{3^{10}\cdot2^7\cdot3^7\cdot2^2}{2^9\cdot5^9\cdot5^8}=\frac{3^{17}\cdot2^9}{2^9\cdot5^{17}}=\frac{3^{17}}{5^{17}}\)

Ta có : \(3^{17}\cdot5^{18}=3^{17}\cdot5^{17}\cdot5=\left(3\cdot5\right)^{17}\cdot5=15^{17}\cdot5\)

\(3^{18}\cdot5^{17}=3\cdot3^{17}\cdot5^{17}=3\cdot\left(3\cdot5\right)^{17}=3\cdot15^{17}\)

\(\text{ Vì }5\cdot15^{17}>3\cdot15^{17}\text{ }\Rightarrow\text{ }3^{17}\cdot5^{18}>3^{18}\cdot5^{17}\text{ }\Rightarrow\text{ }\frac{3^{18}}{5^{18}}< \frac{3^{17}}{5^{17}}\)

15 tháng 9 2019

cảm ơn nha

a: \(=\dfrac{3^6\cdot2^{21}}{5^{18}\cdot7^9\cdot7^3}=\dfrac{3^6\cdot2^{21}}{5^{18}\cdot7^{12}}\)

b: \(=\dfrac{3^{10}\cdot3^7\cdot2^7\cdot2^2}{2^9\cdot5^9\cdot5^8}=\dfrac{3^{17}}{5^{17}}\)

28 tháng 12 2021

\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

4 tháng 1 2022

a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)

b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)

c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)