a)    \({3^x} = 9 \Leftrightarrow {3^x} = {3^2} \Leftrightarrow x = 2\)

\({3^x} = \frac{1}{9} \Leftrightarrow {3^x} = {3^{ - 2}} \Leftrightarrow x =  - 2\)

b)    Có 1 số thực x thỏa mãn: \({3^x} = 5\)

Số năm Trái Đất để sao Hỏa quay quanh mặt trời là:

\(P=d^{\dfrac{3}{2}}=1.52^{1,5}\simeq1,87\left(AU\right)\)

\(a,\sqrt{42}=\sqrt{3\cdot14}>\sqrt{3\cdot12}=6\\ \sqrt[3]{51}=\sqrt[3]{17}< \sqrt[3]{3\cdot72}=6\\ \Rightarrow\sqrt{42}>\sqrt[3]{51}\\ b,16^{\sqrt{3}}=4^{2\sqrt{3}}\\ 18>12\Rightarrow3\sqrt{2}>2\sqrt{3}\Rightarrow4^{3\sqrt{2}}>4^{2\sqrt{3}}\\ \Rightarrow4^{3\sqrt{2}}>16^{\sqrt{3}}\)

\(c,\left(\sqrt{16}\right)^6=16^3=4^6=4^2\cdot4^4=4^2\cdot16^2\\ \left(\sqrt[3]{60}\right)^6=60^2=4^2\cdot15^2\\ 4^2\cdot16^2>4^2\cdot15^2\Rightarrow\sqrt{16}>\sqrt[3]{60}\Rightarrow0,2^{\sqrt{16}}< 0,2^{\sqrt[3]{60}}\)

\(a,1^{1,5}=1;3^{-1}=\dfrac{1}{3};\left(\dfrac{1}{2}\right)^{-2}=4\\ \Rightarrow\dfrac{1}{3}< 1< 4\\ b,2022^0=1;\left(\dfrac{4}{5}\right)^{-1}=\dfrac{5}{4};5^{\dfrac{1}{2}}=\sqrt{5}\simeq2,24\\ \Rightarrow1< \dfrac{5}{4}< \sqrt{5}\)

\(a,a^{\dfrac{1}{3}}\cdot\sqrt{a}=a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{2}}=a^{\dfrac{5}{6}}\\ b,b^{\dfrac{1}{2}}\cdot b^{\dfrac{1}{3}}\cdot\sqrt[6]{b}=b^{\dfrac{1}{2}}\cdot b^{\dfrac{1}{3}}\cdot b^{\dfrac{1}{6}}=b^1\)

\(c,a^{\dfrac{4}{3}}:\sqrt[3]{a}=a^{\dfrac{4}{3}}:a^{\dfrac{1}{3}}=a^{\dfrac{4}{3}-\dfrac{1}{3}}=a\\ d,\sqrt[3]{b}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}}:b^{\dfrac{1}{6}}=b^{\dfrac{1}{3}-\dfrac{1}{6}}=b^{\dfrac{1}{6}}=\sqrt[6]{b}\)

\(a,\left(\dfrac{1}{256}\right)^{-0,75}+\left(\dfrac{1}{27}\right)^{-\dfrac{4}{3}}\\ =256^{\dfrac{3}{4}}+27^{\dfrac{4}{3}}\\ =\sqrt[4]{256^3}+\sqrt[3]{27^4}\\ =145\\ b,\left(\dfrac{1}{49}\right)^{-1,5}-\left(\dfrac{1}{256}\right)^{-\dfrac{2}{3}}\\ =49^{\dfrac{3}{2}}-256^{\dfrac{2}{3}}\\ \simeq343-40,3\\ \simeq302,7\)

\(2\sqrt{3}=\sqrt{12}< \sqrt{18}=3\sqrt{2}\)

=>\(2^{2\sqrt{3}}< 2^{3\sqrt{2}}\)

\(a^m\cdot a^n=a^{m+n}\)

\(a^m:a^n=a^{m-n}\)

\(\left(a^m\right)^n=a^{m\cdot n}\)

\(\left(a\cdot b\right)^m=a^m\cdot b^m\)

\(\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}\)

Vì \(\sqrt{2}>1\)

nên \(10^{\sqrt{2}}>10^1=10\)