Ta có: \(A=\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=\frac{10^{50}-1}{10^{50}-1}+\frac{3}{10^{50}-1}=1+\frac{3}{10^{50}-1}\)

\(B=\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=\frac{10^{50}-3}{10^{50}-3}+\frac{3}{10^{50}-3}=1+\frac{3}{10^{50}-3}\)

Vì \(\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\Rightarrow1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\Rightarrow A< B\)

\(A=\frac{10^{50}+2}{10^{50}+1}=\frac{2}{1}=2\)

\(B=\frac{10^{50}}{10^{50}-3}=\frac{-1}{3}\)

\(\Rightarrow A>B\)

Ta thấy \(10^{50}>10^{50}-3\)

\(\Rightarrow B=\frac{10^{50}}{10^{50}-3}>\frac{10^{50}+2}{10^{50}-3+2}=\frac{10^{50}+2}{10^{50}-1}=A\)

Vậy \(A< B\)

Mình chưa học đến đó nên mình tịt

⇒B=10501050−3 >1050+21050−3+2 =1050+21050−1 =A

Vậy 

C1:A = \(\frac{10^{50}+2}{10^{50}-1}=\frac{10^{50}-1+3}{10^{50}-1}=\frac{10^{50}-1}{10^{50}-1}+\frac{3}{10^{50}-1}\)
= \(1+\frac{3}{10^{50}-1}\)
B = \(\frac{10^{50}}{10^{50}-3}=\frac{10^{50}-3+3}{10^{50}-3}=\frac{10^{50}-3}{10^{50}-3}+\frac{3}{10^{50}-3}\)
= \(1+\frac{3}{10^{50}-3}\)
Vì \(\frac{3}{10^{50}-1}< \frac{3}{10^{50}-3}\)=) \(1+\frac{3}{10^{50}-1}< 1+\frac{3}{10^{50}-3}\)=) \(A< B\)
C2: Áp dụng tính chất : Nếu \(\frac{a}{b}>1\)=) \(\frac{a}{b}>\frac{a+m}{b+m}\)
Vì B > 1 =) B > \(\frac{10^{50}+2}{10^{50}-3+2}=\frac{10^{50}+2}{10^{50}-1}=A\)
(=) B > A

Ta có: \(\dfrac{10^{50}-3}{10^{50}+1}\)<\(\dfrac{10^{50}+1}{10^{50}+1}\)<\(\dfrac{10^{50}+1}{10^{50}-3}\)

=>\(\dfrac{10^{50}-3}{10^{50}+1}\)<​\(\dfrac{10^{50}+1}{10^{50}-3}\)​

vậy (đpcm)

a)

Vì 3<5

\(\Rightarrow3^{30}< 5^{30}\)

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

b)

Ta có

\(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^4\right]^{10}.\left(\frac{1}{2}\right)^{10}\)

\(=\left(\frac{1}{16}\right)^{10}.\left(\frac{1}{2}\right)^{10}\)

Ta có

\(\left(\frac{1}{2}\right)^{10}< 1\)

\(\Leftrightarrow\left(\frac{1}{16}\right)^{10}.\left(\frac{1}{2}\right)^{10}< \left(\frac{1}{16}\right)^{10}\)

\(\Leftrightarrow\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{16}\right)^{10}\)

ta có :\(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)là 2 lũy thừa bậc chẵn nên :\(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)

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

từ trên suy ra (-5)^30<(-3)^50

b) Ta có:\(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2^5}\right)^{10}=\left(\frac{1}{32}\right)^{10}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

ta có :\(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)là 2 lũy thừa bậc chẵn nên :\(\left(-5\right)^{30}=5^{30}=\left(5^3\right)^{10}=125^{10}\)

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

từ trên suy ra (-5)^30<(-3)^50

b) Ta có:\(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2^5}\right)^{10}=\left(\frac{1}{32}\right)^{10}\)

\(\Rightarrow\left(\frac{1}{2}\right)^{50}< \left(\frac{1}{16}\right)^{10}\)

\(\left(\frac{1}{16}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)

Ta có: \(\left(\frac{1}{2}\right)^{50}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)

Do \(\frac{1}{6}>\frac{1}{32}\Rightarrow\left(\frac{1}{6}\right)^{10}>\left(\frac{1}{32}\right)^{10}\)

Vậy \(\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

a) \(10^{20}\) và \(9^{10}\)

Vì 10 > 9 ; 20 > 10

nên \(10^{20}>9^{10}\)

Vậy \(10^{20}>9^{10}\)

b) \(\left(-5\right)^{30}\) và \(\left(-3\right)^{50}\)

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

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

Vì 243 > 125 nên \(125^{10}< 243^{10}\)

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

c) \(64^8\) và \(16^{12}\)

Ta có: \(64^8=\left(4^3\right)^8=4^{24}\)

          \(16^{12}=\left(4^2\right)^{12}=4^{24}\)

Vậy \(64^8=16^{12}\left(=4^{24}\right)\)

d) \(\left(\frac{1}{6}\right)^{10}\) và \(\left(\frac{1}{2}\right)^{50}\)

Ta có: \(\left(\frac{1}{6}\right)^{10}=\left[\left(\frac{1}{2}\right)^4\right]^{10}=\left(\frac{1}{2}\right)^{40}\)

Vì 40 < 50 nên \(\left(\frac{1}{2}\right)^{40}< \left(\frac{1}{2}\right)^{50}\)

Vậy \(\left(\frac{1}{16}\right)^{10}< \left(\frac{1}{2}\right)^{50}\)

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 \(2^{x+1}\cdot3^{y}=12^{x}\)

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

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