`[(0,6)^{3}.(0,6)^{8}].[(0,6)^{7}.(0,6)^{2}]`

`=(0,6)^{3+8}.(0,6)^{7+2}`

`=(0,6)^{11}.(0,6)^9`

`=(0,6)^{11+9}`

`=(0,6)^{20}`

\((\dfrac{7}{8})^{5}:(\dfrac{7}{8})^{4}\)

\(=(\dfrac{7}{8})^{5-4}\)

\(=\dfrac{7}{8}\)

`[(3/7)^4:(3/7)^5]. 7/8`

`=(3/7)^{4-5} .7/8`

`=(3/7)^{-1}. 7/8`

`=1/[3/7] . 7/8`

`=7/3 . 7/8`

`=49/24`

A = 3 - \(\sqrt{x-6}\)

đk x ≥ 6 ,  A = 3 - \(\sqrt{x-6}\)

\(\sqrt{x-6}\) ≥ 0 ⇔ - \(\sqrt{x-6}\) ≤ 0 ⇔ 3- \(\sqrt{x-6}\) ≤ 3 ⇔ A(max)= 3 ⇔ x = 6

\(\dfrac{\left(0,04\right)^5}{\left(0,2\right)^6}=\dfrac{\left(\dfrac{1}{25}\right)^5}{\left(\dfrac{1}{5}\right)^6}=\dfrac{\left(\left(\dfrac{1}{5}\right)^2\right)^5}{\left(\dfrac{1}{5}\right)^6}=\dfrac{\left(\dfrac{1}{5}\right)^{10}}{\left(\dfrac{1}{5}\right)^6}=\left(\dfrac{1}{5}\right)^4=\dfrac{1}{625}\)

\(\dfrac{2^7.9^3}{8^2.2^5.3^5}=\dfrac{2^7.\left(3^2\right)^3}{\left(2^3\right)^2.2^5.3^5}=\dfrac{2^7.3^6}{2^6.2^5.3^5}=\dfrac{2^7.3^6}{2^{11}.3^5}=\dfrac{3}{2^4}=\dfrac{3}{16}\)

\(\dfrac{2^3.3^3.3.2^2.3^2+3^3}{-13}=\dfrac{2^3.2^2.3^3.3.3^2+3^3}{-13}=\dfrac{2^5.3^6+3^3}{-13}=\dfrac{3^3.\left(2^5.3^3+1\right)}{-13}=\dfrac{3^3.\left(864+1\right)}{-13}=\dfrac{3^3.865}{-13}=\dfrac{23355}{-13}\)

A = \(\dfrac{27^{10}+9^5}{9^{13}+27^2}\)

A = 81

A = \(\dfrac{27^{10}+9^5}{9^{13}+27^2}\)

A= \(\dfrac{27^8}{^{ }9^8}\)

Tìm số tận cùng của các số sau

\(87^{32}=\left(87^2\right)^{16}=7569^{16}=.......1\)

Vậy chữ số tận cùng của 87³² là 1

\(58^{33}=2^{33}.29^{33}=2^{11}.4^{11}.29^{33}=2048\times...4\times...9=.....8\)

Vậy chữ số tập cùng của 58³³ là 8

$x\; \in BC(39;65;91)$

Ta có:

$39=3.13$

$65=13.5$

$91=7.13$

$\Rightarrow BCN$$N(39;65;91)=3.5.7.13=1365$

$\Rightarrow BC(39;65;91)=\{0;1365;2730;...\}$

Mà $4000<x<6000$

$\Rightarrow x=4095;5460.$