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

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

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

1.a)A = (1 - 1/3)(1-2/5)...(1-5/5)....(1-9/5)

      =(1-1/3)....0.....(1-9/5)

      =0

     =>đpcm.

b)ta xét:

1/2² = 1/2x2 < 1/1x2

.............

1/8² = 1/8x8 <1/7x8

=>B < 1/1x2 + 1/2x3 ... + 1 + 1/7x8

<=> B <1 - 1/2 + 1/2  - 1/3  + ... + 1/7 - 1/8

<=> B < 1 - 1/8 = 7/8 < 1

=> B < 1 => đpcm

2.a) Đặt m = 2007(2006+2007) = 2006(2006 + 2007) + (2006+2007)

      Đặt n = 2006(2007+2008) = 2006(2006+2007) + (2006 + 2006)

Ta thấy : (2006+2007) > (2006 + 2006) => m > n , áp dụng công thức "a.d > c.d <=> a/b > b/d (a,c thuộc Z// b,d thuộc N)

=> A > B

   b)ta có: D = 196 + 197/197 + 198 = (196/197+198) + (197/197+198) < 196/197 + 197/198 = C

=> C > D

c)gọi 20¹⁰ là a

ta thấy : (a + 1)(a-3) = (a - 1)(a - 3) + 2(a - 3) < (a - 1)(a - 3) + 2(a - 1) = (a - 1)(a - 1)

áp dụng: ad > bc <=> a/b > c/d ( a,b,c,d thuộc Z// b,d > 0)

=> E > F

b.Ta có :  = (111.3)^111.4 = ( 111⁴ . 3⁴ )¹¹¹= 

            = ( 111 . 4 )^111.3 = ( 111³.4³)111 = 

Vì (111⁴.81)¹¹¹ > ( 111³.64 )¹¹¹ => 333⁴⁴⁴ > 444³³³

a. 10³⁰ = ( 10³ )¹⁰=1000¹⁰

2¹⁰⁰ = ( 2¹⁰ )¹⁰=1024¹⁰

Vì 1000¹⁰<1024¹⁰ nên 10³⁰<2¹⁰⁰

a,5mũ 36=(5mũ3)mũ12=125 mũ12

11^24=(11^2)12=121^12

vì 121<125 nên 5^36>11^24

cảm ơn nha

a,

15^12=(3*5)^12=3^12*5^12

81^3*125^5=(3^4)^3*(5^3)^5=3^12*5^15

Vì 12<15 suy ra 5^12<5^15

Suy ra 3^12*5^12<3^12*5^15

\(a.81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{12}.5^{15}=3^{12}.5^{12}.5^3=\left(3.5\right)^{12}.5^3=15^{12}.5^3>15^{12}\)

\(b.4^{20}.81^{12}=\left(2^2\right)^{20}.\left(9^2\right)^{12}=2^{40}.9^{24}=2^{20}.2^{20}.9^{20}.9^4=\left(2.9\right)^{20}.2^{20}.9^4=18^{20}.2^{20}.9^4>18^{20}\)

\(c.73^{75}=\left(73^3\right)^{25}=389017^{25}\)

\(107^{50}=107^{2.50}=\left(107^2\right)^{25}=11449^{25}\)

Vì \(389017^{25}>11449^{25}\Rightarrow73^{75}>107^{50}\)