\(5A=\dfrac{5^{2022}+5}{5^{2022}+1}=1+\dfrac{4}{5^{2022}+1}\)

Sửa đề: \(B=\dfrac{5^{2020}+1}{5^{2021}+1}\)

=>\(5B=\dfrac{5^{2021}+5}{5^{2021}+1}=1+\dfrac{4}{5^{2021}+1}\)

5^2022>5^2021

=>5^2022+1>5^2021+1

=>5A<5B

=>A<B

Ta có: \(A=\frac{5^{2020}+1}{5^{2020}+1}=1\)

\(B=\frac{5^{2019}+1}{5^{2020}+1}< 1\)

=> B < A

(◍•ᴗ•◍)

Đầu tiên chúng ta sẽ so sánh như sau

5^2010 và 5^2009

vì 2010>2009 nên 5^2010>5^200 (1)

1/5^2011+1 và 1/5^2010+1

vì 2011+1=2012

   2010+1=2011

mà 2012>2011 nên 1/5^2011+1>1/5^2010+1 (2)

Từ 1 và 2 ta có thể suy ra A>B

Vậy A>B

ta có 2010 >2009 suy ra 5^2010 >5^2009 suy ra 5^2010 + 1>5^2009 +1                                               (1)

         2011>2010 suy ra 5^2011 >5^2010 suy ra 1/5^2011<1/5^2010 suy ra 1/5^2011 +1 <1/5^2010 + 1  (2)

từ (1) và (2) => A=B

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

(3/8)⁵= 3⁵/(2³)⁵=243/2¹⁵>243/3¹⁵>125/3¹⁵=5³/(3⁵)³=(5/3⁵)³=(5/243)³

suy ra (3/8)⁵>(5/243)³