Câu a)
\(\hept{\begin{cases}A=3,4\cdot10^{-8}=3,4\cdot\frac{1}{10^8}=\frac{3,4}{10^8}\\B=34\cdot10^{-9}=34\cdot\frac{1}{10^9}=\frac{3,4\cdot10}{10^8\cdot10}=\frac{3,4}{10^8}\end{cases}}\)
\(\Rightarrow\frac{A}{B}=1\)

Câu b)
\(\hept{\begin{cases}A=10^{-4}+10^{-3}+10^{-2}=\frac{1}{10^4}+\frac{1}{10^3}+\frac{1}{10^2}=\frac{10^5+10^6+10^7}{10^9}\\B=10^{-9}=\frac{1}{10^9}\end{cases}}\)
\(\Rightarrow\frac{A}{B}=10^5+10^6+10^7=11100000\)

a ) A gấp B 10 lần

\(\) 10-1+18-17+15-12=?

Ta có:\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}\)

\(\Rightarrow A=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}\)

\(\Rightarrow A=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}\)

\(\Rightarrow B=1+\frac{3}{10^8-3}\)

Vì \(\frac{3}{10^8-1}>\frac{3}{10^8-3}\Rightarrow A>B\)

mk nghĩ là a>b

a>b 100% đúng^_^

thánh nó trl

\(\text{So sánh : }\)

\(A=2017^{10}+2017^9=1,1149984e33\)

\(B=2018^{10}=1,11998349e33\)

\(\text{Vì : }1,1149984e33< 1,11998349e33\text{ nên }A< B\)

a, 8¹⁴ và 9²¹= 8^2.7và 9^3.7

                  = (8²)⁷ và (9³)⁷

                  = 16⁷ và 21⁷

do 16<21 nên 16⁷<21⁷ hay 8¹⁴<9²¹

b,5⁴⁰ và 620¹⁰= 5^4.10 và 620¹⁰

                           = (5⁴)¹⁰ và 620¹⁰

                    = 20¹⁰ và 620¹⁰

do 20<620 nên 20¹⁰< 620¹⁰ hay 5⁴⁰<620¹⁰

8 mũ 2 chứ ko phải 8 nhân 2 nha bạn

a: \(\dfrac{A}{B}=\dfrac{3.4\cdot10^{-8}}{3.4\cdot10^{-9}}=\dfrac{1}{10^8}:\dfrac{1}{10^9}=10\)

=>A gấp 10 lần B

b: \(\dfrac{A}{B}=\dfrac{10^{-4}+10^{-3}+10^{-2}}{10^{-9}}=\dfrac{10^{-2}\cdot\left(10^{-2}+10^{-1}+1\right)}{10^{-9}}=\dfrac{1}{10^7}\cdot\dfrac{111}{100}=\dfrac{111}{10^9}\)

=>A bằng 111/10⁹ lần B

Bài 1:

a: \(10^{10}=\left(2\cdot5\right)^{10}=2^{10}\cdot5^{10}=2^9\cdot5^{10}\cdot2\)

\(48\cdot50^5=2^4\cdot3\cdot\left(2\cdot5^2\right)^5=2^4\cdot3\cdot2^5\cdot5^{10}=2^9\cdot5^{10}\cdot3\)

mà 2<3

nên \(10^{10}<48\cdot50^5\)

b: \(1990^{10}+1990^9=1990^9\left(1990+1\right)=1990^9\cdot1991\)

\(1991^{10}=1991^9\cdot1991\)

mà 1990<1991

nên \(1990^{10}+1990^9<1991^{10}\)

c: \(107^{50}<108^{50}=\left(2^2\cdot3^3\right)^{50}=2^{100}\cdot3^{150}\)

\(73^{75}>72^{75}=\left(2^3\cdot3^2\right)^{75}=2^{225}\cdot3^{150}\)

mà \(2^{225}\cdot3^{150}>2^{100}\cdot3^{150}=108^{50}>107^{50}\)

nên \(73^{75}>107^{50}\)

d: \(2^{91}=\left(2^{13}\right)^7=8192^7\)

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

mà 8192>3125

nên \(2^{91}>5^{35}\)

e: \(A=72^{45}-72^{44}=72^{44}\left(72-1\right)=72^{44}\cdot71\)

\(B=72^{44}-72^{43}=72^{43}\left(72-1\right)=72^{43}\cdot71\)

mà 44>43

nên A>B

Bài 2:

a:

ĐKXĐ: x<>2023

\(\frac{x-2023}{4}=\frac{1}{x-2023}\)

=>\(\left(x-2023\right)\left(x-2023\right)=4\cdot1\)

=>\(\left(x-2023\right)^2=4\)

=>\(\left[\begin{array}{l}x-2023=2\\ x-2023=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2+2023=2025\left(nhận\right)\\ x=-2+2023=2021\left(nhận\right)\end{array}\right.\)

b: \(\left(2x+1\right)^4=\left(2x+1\right)^6\)

=>\(\left(2x+1\right)^6-\left(2x+1\right)^4=0\)

=>\(\left(2x+1\right)^4\cdot\left\lbrack\left(2x+1\right)^2-1\right\rbrack=0\)

=>\(\left(2x+1\right)^4\cdot\left(2x+1-1\right)\left(2x+1+1\right)=0\)

=>\(2x\left(2x+1\right)^4\cdot\left(2x+2\right)=0\)

=>\(\left[\begin{array}{l}2x=0\\ 2x+1=0\\ 2x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac12\\ x=-1\end{array}\right.\)

c: \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

=>\(\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)

=>\(\left(3x-1\right)^{10}\cdot\left\lbrack\left(3x-1\right)^{10}-1\right\rbrack=0\)

=>\(\left[\begin{array}{l}\left(3x-1\right)^{10}=0\\ \left(3x-1\right)^{10}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x-1=0\\ \left(3x-1\right)^{10}=1\end{array}\right.\)

=>\(\left[\begin{array}{l}3x-1=0\\ 3x-1=1\\ 3x-1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=\frac23\\ x=0\end{array}\right.\)

d: Sửa đề \(2^{x+1}\cdot3^{y}=12^{x}\)

=>\(2^{x+1}\cdot3^{y}=\left(2^2\cdot3\right)^{x}=2^{2x}\cdot3^{x}\)

=>\(\begin{cases}2x=x+1\\ y=x\end{cases}\Rightarrow\begin{cases}x=1\\ y=x=1\end{cases}\)

a,>

b,>

c,<

a, 19²⁰ > 9⁸

b, 5⁴⁰ < 620¹⁰

c, Ta có: \(2^{161}=2^{7.23}=\left(2^7\right)^{23}=128^{23}\)

=> 128²³ > 13⁴⁰ hay 2¹⁶¹ > 13⁴⁰