Đáp án:

$\begin{array}{l}
42)\\
a){\left( {\dfrac{2}{5}} \right)^x} = \dfrac{4}{{25}} = {\left( {\dfrac{2}{5}} \right)^2}\\
 \Rightarrow x = 2\\
b){\left( {\dfrac{2}{3}} \right)^x} = \dfrac{4}{9} = {\left( {\dfrac{2}{3}} \right)^2}\\
 \Rightarrow x = 2\\
c){\left( {\dfrac{2}{5}} \right)^x} = {\left( {\dfrac{4}{{25}}} \right)^2} = {\left( {\dfrac{2}{5}} \right)^{2.2}} = {\left( {\dfrac{2}{5}} \right)^4}\\
 \Rightarrow x = 4\\
d){\left( {\dfrac{2}{3}} \right)^x} = {\left( {\dfrac{4}{9}} \right)^4} = {\left( {\dfrac{2}{3}} \right)^{2.4}} = {\left( {\dfrac{2}{3}} \right)^8}\\
 \Rightarrow x = 8\\
e){\left( {\dfrac{3}{5}} \right)^x} = {\left( {\dfrac{9}{{25}}} \right)^3} = {\left( {\dfrac{3}{5}} \right)^6}\\
 \Rightarrow x = 6\\
f){\left( {\dfrac{{ - 3}}{2}} \right)^x} = \dfrac{9}{4} = {\left( {\dfrac{{ - 3}}{2}} \right)^2}\\
 \Rightarrow x = 2\\
43)\\
a)\dfrac{{32}}{{{2^n}}} = 2\\
 \Rightarrow \dfrac{{{2^5}}}{{{2^n}}} = 2\\
 \Rightarrow {2^{5 - n}} = {2^1}\\
 \Rightarrow 5 - n = 1\\
 \Rightarrow n = 4\\
b)\dfrac{{{8^n}}}{{{2^n}}} = 16\\
 \Rightarrow \dfrac{{{2^{3n}}}}{{{2^n}}} = {2^4}\\
 \Rightarrow {2^{3n - n}} = {2^4}\\
 \Rightarrow 2n = 4\\
 \Rightarrow n = 2\\
c)\dfrac{{{{33}^{2n}}}}{{{{11}^{2n}}}} = 81\\
 \Rightarrow {\left( {\dfrac{{33}}{{11}}} \right)^{2n}} = {3^4}\\
 \Rightarrow {3^{2n}} = {3^4}\\
 \Rightarrow 2n = 4\\
 \Rightarrow n = 2\\
d)\dfrac{{54}}{{{3^{n - 2}}}} = 2\\
 \Rightarrow {3^{n - 2}} = \dfrac{{54}}{2} = 27 = {3^3}\\
 \Rightarrow n - 2 = 3\\
 \Rightarrow n = 5\\
e)\dfrac{{175}}{{{5^{n + 1}}}} = 7\\
 \Rightarrow {5^{n + 1}} = \dfrac{{175}}{7} = 25 = {5^2}\\
 \Rightarrow n + 1 = 2\\
 \Rightarrow n = 1\\
f)\dfrac{{135}}{{{3^{n - 2}}}} = 5\\
 \Rightarrow {3^{n - 2}} = 27 = {3^3}\\
 \Rightarrow n = 5\\
g)\dfrac{{64}}{{{2^{n + 1}}}} = 8\\
 \Rightarrow {2^{n + 1}} = 8 = {2^3}\\
 \Rightarrow n = 2\\
h)\dfrac{{{3^4}}}{{{3^n}}} = {3^{20}}\\
 \Rightarrow {3^{4 - n}} = {3^{20}}\\
 \Rightarrow 4 - n = 20\\
 \Rightarrow n =  - 16\\
i)1 - n = 10\\
 \Rightarrow n =  - 9
\end{array}$

j) ko nhìn rõ đề bài

$\begin{array}{l}
k)\dfrac{{{2^3}}}{{{2^n}}} = {4^3}\\
 \Rightarrow {2^{3 - n}} = {2^6}\\
 \Rightarrow 3 - n = 6\\
 \Rightarrow n =  - 3\\
k){3^n} = \dfrac{{{9^8}}}{{{{27}^3}{{.81}^2}}} = \dfrac{{{3^{2.8}}}}{{{3^{3.3}}{{.3}^{4.2}}}}\\
 \Rightarrow {3^n} = {3^{16 - 9 - 8}} = {3^{ - 1}}\\
 \Rightarrow n =  - 1
\end{array}$

Đáp án:

 Bài 42:

a) x=2

b) x=2

c) x=2

d) x=2

e) x=2

f) x=2

Giải thích các bước giải:

 Trong kiến thức cần nhớ nha !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!