A và B có phần mẫu số bằng nhau mà tử A có 10^2023 lớn hơn B có 10^2022 => A > B

10^2023>10^2022

=>10^2023+5>10^2022+5

=>A>B

\(\dfrac{1}{10}A=\dfrac{10^{2023}+5}{10^{2023}+50}=1-\dfrac{45}{10^{2023}+50}\)

\(\dfrac{1}{10}B=\dfrac{10^{2022}+5}{10^{2022}+50}=1-\dfrac{45}{10^{2022}+50}\)

10^2023+50>10^2022+50

=>-45/10^2023+50<-45/10^2020+50

=>1/10A<1/10B

=>A<B

Bài 1: 

a) 0²⁰⁰² < 0²⁰²³

b) 2022⁰ = 2023⁰

c) 54⁹ < 55¹⁰

d) ( 4 + 5 )³ > 4² + 5²

đ) 9² - 3² > ( 9 - 3 )²

Bài 2:

a) 3² x 4³ - 3² + 333

= 9 x 64 - 9 + 333

= 576 - 9 + 333

= 567 + 333

= 900

b) 5 x 4³ + 24 x 5 + 41⁰

= 5 x 64 + 24 x 5 + 1

= 5 x ( 64 + 24 ) + 1

= 5 x 88 + 1

= 440 + 1

= 441

c) 2³ x 4² + 3² x 5 - 40 x 1²⁰²³

= 8 x 16 + 9 x 5 - 40 x 1

= 128 + 45 - 40

= 133

Bài 1 :

a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)

b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)

c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)

d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)

đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)

Bài 2 :

a) \(3^2.4^3-3^2+333=3^2\left(4^3-1\right)+9.37=9.63+9.37=9\left(63+37\right)=9.100=900\)

b) \(5.4^3+24.5+41^0=20.4^2+20.6+1=20\left(16+6\right)+1=20.22+1=441\)

c) \(2^3.4^2+3^2.5-40.1^{2023}=8.16+9.5-40.1=128+45-40=128+5=133\)

\(a)\dfrac{7}{8}=\dfrac{7\times9}{8\times9}=\dfrac{63}{72}\)

\(\dfrac{3}{9}=\dfrac{3\times8}{9\times8}=\dfrac{24}{72}\)

Do : \(\dfrac{63}{72}>\dfrac{24}{72}\) nên \(\dfrac{7}{8}>\dfrac{3}{9}\)

Không thì bạn có thể rút gọn 3/9 đi làm cho nó gọn ạ.

\(b)\) Ta thấy : \(\dfrac{2023}{2021}>1\) ( vì tử lớn hơn mẫu )

                   \(\dfrac{2021}{2022}< 1\) ( vì tử bé hơn mẫu )

Do đó : \(\dfrac{2023}{2021}>\dfrac{2021}{2022}\)

\(c)\dfrac{5}{6}=\dfrac{5\times7}{6\times7}=\dfrac{35}{42}\)

\(\dfrac{6}{7}=\dfrac{6\times6}{7\times6}=\dfrac{36}{42}\)

Do : \(\dfrac{36}{42}>\dfrac{35}{42}\)  nên \(\dfrac{6}{7}>\dfrac{5}{6}\)

không câu SP nhé

guihopijp['j'OPHGPG

Áp dụng tính chất : Nếu \(\dfrac{a}{b}< 1\) thì \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\) ( a; b; n ϵ N , b; n ≠ 0 )

Ta có \(\dfrac{2023^{31}+5}{2023^{32}+5}< 1\)

⇒ \(B=\dfrac{2023^{31}+5}{2023^{32}+5}< \dfrac{2023^{31}+5+2018}{2023^{32}+5+2018}=\dfrac{2023^{31}+2023}{2023^{32}+2023}=\dfrac{2023\left(2023^{30}+1\right)}{2023\left(2023^{31}+1\right)}=\dfrac{2023^{30}+1}{2023^{31}+1}=A\)Vậy A > B

Ta có 2023A = \(\dfrac{2023.\left(2023^{30}+5\right)}{2023^{31}+5}=\dfrac{2023^{31}+5.2023}{2023^{31}+5}\)

\(=1+\dfrac{2022.5}{2023^{31}+5}\)

Lại có 2023B = \(\dfrac{2023.\left(2023^{31}+5\right)}{2023^{32}+5}=\dfrac{2023^{32}+2023.5}{2023^{32}+5}\)

\(=1+\dfrac{2022.5}{2023^{32}+5}\)

Dễ thấy 2023³¹ + 5 < 2023³² + 5

\(\Leftrightarrow\dfrac{2022.5}{2023^{31}+5}>\dfrac{2022.5}{2023^{32}+5}\)

\(\Leftrightarrow1+\dfrac{2022.5}{2023^{31}+5}>1+\dfrac{2022.5}{2023^{32}>5}\)

\(\Leftrightarrow2023A>2023B\Leftrightarrow A>B\)

a: \(B=\dfrac{154}{155+156}+\dfrac{155}{155+156}\)

\(\dfrac{154}{155}>\dfrac{154}{155+156}\)

\(\dfrac{155}{156}>\dfrac{155}{155+156}\)

=>154/155+155/156>(154+155)/(155+156)

=>A>B

b: \(C=\dfrac{2021+2022+2023}{2022+2023+2024}=\dfrac{2021}{6069}+\dfrac{2022}{6069}+\dfrac{2023}{6069}\)

2021/2022>2021/6069

2022/2023>2022/2069

2023/2024>2023/6069

=>D>C

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