a) Ta có: \(10^{20}=\left(10^2\right)^{10}=100^{10}\)

Mà \(100^{10}>19^{10}\)

\(\Rightarrow10^{20}>19^{10}\)

b) Ta có: \(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)

\(\left(-3\right)^{50}=3^{50}=\left(3^5\right)^{10}=243^{10}\)

Mà: \(125^{10}< 243^{10}\)

\(\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)

c) Ta có: \(64^8=\left(2^6\right)^8=2^{48}\)

\(16^{12}=\left(2^4\right)^{12}=2^{48}\)

Mà: \(2^{48}=2^{48}\)

\(\Rightarrow64^8=16^{12}\)

a) 10²⁰và 19¹⁰

Ta có: 10²⁰= (10²)¹⁰ và 19¹⁰

                   = 100¹⁰ và 19¹⁰

Vì 100¹⁰>19¹⁰ => 10²⁰>19¹⁰

b) (-5)³⁰ và (-3)⁵⁰

Ta có:

 (-5)30= [(-5)³]¹⁰=(-125)¹⁰ và  (-3)⁵⁰=[(-3)⁵]¹⁰=(-243)¹⁰

Vì -125¹⁰>-243¹⁰ Nên (-5)³⁰>(-3)⁵⁰

 c) 64⁸ và 16¹²  

= (4³)⁸và  (4²)¹²

= 4²⁴ và 4²⁴

=> 64⁸ = 16¹²

a ] Ta có : 10\(^{20}\) = 10\(^{^{ }2.10}\) = [ 10\(^2\) ]\(^{10}\) = 100\(^{10}\) 

Vì 100\(^{10}\) > 19\(^{10}\) Nên => 10\(^{20}\) > 19\(^{10}\)

b ] Ta có : [ -5 ]\(^{30}\) = [ -5 ]\(^{3.10}\) = [ -5\(^3\) ]\(^{10}\) = [ -125 ]\(^{10}\) 

                [ -3 ]\(^{50}\) = [ -3 ]\(^{5.10}\) = [ -3\(^5\) ]\(^{10}\) = [ -243 ]\(^{10}\)

Vì [ -125 ]\(^{10}\) < [ -243 ]\(^{10}\) Nên => [ -5 ]\(^{30}\) < [ -3 ]\(^{50}\)

c ] Ta có : 64\(^8\) = 64\(^{2.4}\) = [ 64\(^2\) ]\(^4\) = 4196\(^4\) 

                16\(^{12}\) = 16\(^{2.6}\) = [ 16\(^2\) ]\(^6\) = 4096\(^4\) 

Vì 4196\(^4\) > 4096\(^4\) Nên => 64\(^8\) > 16\(^{12}\)

1.

b) \(3^x+3^{x+2}=2430\)

\(\Rightarrow3^x.1+3^x.3^2=2430\)

\(\Rightarrow3^x.\left(1+3^2\right)=2430\)

\(\Rightarrow3^x.10=2430\)

\(\Rightarrow3^x=2430:10\)

\(\Rightarrow3^x=243\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)

Chúc bạn học tốt!

3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

\(3.-12+\left(2x-9\right)+x=0.\)

\(\Leftrightarrow-12+2x-9+x=0.\Leftrightarrow3x=21.\Leftrightarrow x=7.\)

Vậy \(x=7.\)

\(4.11+\left(15-x\right)=1.\Leftrightarrow11+15-x=1.\Leftrightarrow26-x=1.\Leftrightarrow x=25.\)

Vậy \(x=25.\)

\(5.4-\left(27-3\right)=x-\left(13-4\right).\Leftrightarrow4-24=x-9.\Leftrightarrow-20=x-9.\Leftrightarrow x=-11.\)

Vậy \(x=-11.\)

\(6.8-\left(x-10\right)=23-\left(-4+12\right).\Leftrightarrow8-x+10=23-8.\Leftrightarrow18-x=15.\Leftrightarrow x=3.\)

Vậy \(x=3.\)

\(7.105-5\left(10-5x\right)=-20.\Leftrightarrow105-50+25x=-20.\Leftrightarrow25x=-75.\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

\(8.\left(x-1\right)\left(8-2x\right)\left(3x+123\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\8-2x=0.\\3x+123=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=4.\\x=-41.\end{matrix}\right.\)

Vậy \(x\in\left\{1;4;-41\right\}.\)

\(9.\left(x^2-25\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+10\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0.\\x+5=0.\\x+10=0.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5.\\x=-5.\\x=-10.\end{matrix}\right.\)

Vậy \(x\in\left\{5;-5;-10\right\}.\)

\(10.x\left(x^2+5\right)=0.\Leftrightarrow x=0.\)

1.

a. Ta có: \(A=2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)

\(B=3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)

Mà \(8^{100}< 9^{100}\)

\(\Rightarrow A< B\)

b. Ta có: \(A=2^{332}< 2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)

\(B=3^{223}>3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)

Mà \(8^{111}< 9^{111}\)

\(\Rightarrow A< B\)

c. Ta có: \(A=2^{91}=2^{13.7}=\left(2^{13}\right)^7=8192^7\)

\(B=5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)

Mà \(8192^7>3125^7\)

\(\Rightarrow A>B\)

Câu 2: 

a: =>(x-6)(x-7)=0

=>x=6 hoặc x=7

b: =>\(x^8\left(x^2-25\right)=0\)

\(\Leftrightarrow x^8\left(x-5\right)\left(x+5\right)=0\)

hay \(x\in\left\{0;5;-5\right\}\)

a)

\(\frac{x}{18}=\frac{y}{15},x-y=-30\)

\(\frac{x}{18}=\frac{y}{15}\)

\(\frac{x}{18}-\frac{y}{15}=0\)

\(-\frac{6y-5x}{90}=0\)

\(6y-5x=0\)

\(x-y=-30\)

\(-\left(y-x-30\right)=0\)

\(y-x-30=0\)

\(\Rightarrow x=-180;y=-150\)

dẫn tui bài y:9-8=15 với

a) x - \(\frac{1}{8}\)= \(\frac{7}{3}\)x \(\frac{21}{4}\)

   x - \(\frac{1}{8}\)= \(\frac{49}{4}\)

             x = \(\frac{49}{4}\)+ \(\frac{1}{8}\)

             x = 98

b) x : \(\frac{3}{2}\) = \(\frac{1}{4}\)

               x = \(\frac{1}{4}\)x \(\frac{3}{2}\)

              x = \(\frac{3}{8}\)

c) \(\frac{103}{10}\) - x = \(\frac{4}{5}\)

                    x = \(\frac{103}{10}\)- \(\frac{4}{5}\)

                    x = \(\frac{19}{5}\)

d) x = 51 x \(\frac{4}{17}\)

    x = \(12\)

Vậy chị kết luận ngắn gọn là:

a)=98

b)=3/8

c)=19/5

d)=12