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

\(M=\sqrt{\left(\frac{1}{25}\right)^{\left(-\frac{3}{2}\right)}-\left(\frac{1}{8}\right)^{\left(-\frac{2}{3}\right)}}=\sqrt{\left(5^{-2}\right)^{-\frac{3}{2}}-\left(2^{-3}\right)^{-\frac{2}{3}}}=\sqrt{5^3-2^2}=\sqrt{121}=11\)

a) Ta có: \(\left(0.125\right)^m\cdot24^m=243\)

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

hay m=5

a.

\(\left(\frac{1}{3}\right)^{2x-1}=243\)

\(\left(\frac{1}{3}\right)^{2x-1}=\left(\frac{1}{3}\right)^{-5}\)

\(2x-1=-5\)

\(2x=-5+1\)

\(2x=-4\)

\(x=-\frac{4}{2}\)

\(x=-2\)

b.

\(\left(0,125\right)^{x+1}=64\)

\(\left(0,125\right)^{x+1}=\left(0,125\right)^{-2}\)

\(x+1=-2\)

\(x=-2-1\)

\(x=-3\)

a) c1:

(8 x 24) : 4

=8 x (24 : 4)

=8 x 6

=48

c2:

(8 x 24) : 4

=192 : 4

=48

b) c1

(15 x 24) : 6

=360 : 6

=60

c2:

(15 x 24) : 6

=15 x (24 : 6)

=15 x 4

=60

 a ) ( 8 . 24 ) : 4

 = 192 : 4

= 48

Cách 2 :

  ( 8 . 24 ) : 4

= ( 4 . 2 . 24 ) : 4

= 2 . 24 

= 48 

b ) Tương tự 

60 day

\(\left(\dfrac{1}{8}\right)^m\cdot24^m=243\)

\(\Rightarrow\left(\dfrac{1}{8}\cdot24\right)^m=243\)
\(\Rightarrow3^m=243=3^5\)
\(\Rightarrow m=5\)

(0,125)^m . 24^m = 243

(0,125 . 24)^m =243

3^m =3⁵

=>m = 5

\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^{x+2}=\dfrac{10}{243}\)

\(\left(\dfrac{1}{3}\right)^x+\left(\dfrac{1}{3}\right)^x.\left(\dfrac{1}{3}\right)^2=\dfrac{10}{243}\)

\(\left(\dfrac{1}{3}\right)^x.\left[1+\left(\dfrac{1}{3}\right)^2\right]=\dfrac{10}{243}\)

\(\left(\dfrac{1}{3}\right)^x.\dfrac{10}{9}=\dfrac{10}{243}\)

\(\left(\dfrac{1}{3}\right)^x=\dfrac{10}{243}:\dfrac{10}{9}\)

\(\left(\dfrac{1}{3}\right)^x=\dfrac{1}{27}\)

\(\left(\dfrac{1}{3}\right)^x=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x=3\)

\(=\dfrac{3^5}{-\left(0.003\right)^3\cdot0.09}=-10^{11}\)