a,\(\left(\dfrac{1}{3}\right)^{2n-1}=243\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^{2n-1}=\left(\dfrac{1}{3}\right)^{-5}\)
Vì \(\dfrac{1}{3}\ne\pm1;\dfrac{1}{3}\ne0\) nên
\(2n-1=-5\Rightarrow2n=-4\Rightarrow n=-2\)
Vậy........
b, \(\left(0,125\right)^{n+1}=64\)
\(\Rightarrow\left(\dfrac{1}{8}\right)^{n+1}=\left(\dfrac{1}{8}\right)^{-2}\)
Vì \(\dfrac{1}{8}\ne\pm1;\dfrac{1}{8}\ne0\) nên
\(n+1=-2\Rightarrow n=-3\)
Vậy..........
Chúc bạn học tốt!!!
a) Ta có: \(\left(\dfrac{1}{3}\right)^{2n-1}=243\Rightarrow\left(3^{-1}\right)^{2n-1}=3^5\)
\(\Rightarrow3^{-2n+1}=3^5\)
Suy ra: \(-2n+1=5\Rightarrow-2n=4\Rightarrow n=-2\)
b) \(\left(0,125\right)^{n+1}=64\Rightarrow\left(\dfrac{1}{8}\right)^{n+1}=8^2\)
\(\Rightarrow\left(8^{-1}\right)^{n+1}=8^2\)
\(\Rightarrow8^{-n-1}=8^2\)
Suy ra: \(-n-1=2\Rightarrow n=-3\)
Hok tốt
Cảm ơn hai bạn nha mk có thâm 3 câu hỏi muốn nhờ nữa này
Toshiro KiyoshiSongoku
Toshiro KiyoshiSongoku
Link nè: Câu hỏi của Phát
