Ta có : Q=\(\frac{1010+1011+1012}{1011+1012+1013}\)=\(\frac{1010}{1011+1012+1013}+\frac{1011}{1011+1012+1013}+\frac{1012}{1011+1012+1013}\)

Vì1010/1011>1010/1011+1012+1013

    1011/1012>1011/1011+1012+1013

    1012/1013>1012/1011+1012+1013

    =>P>Q

b) Ta có: \(A=\dfrac{1012+1}{1013+1}\)

\(\Leftrightarrow A-1=\dfrac{1012+1-1013-1}{1013+1}\)

\(\Leftrightarrow A-1=\dfrac{-1}{1013+1}\)

Ta có: \(B=\dfrac{1011+1}{1012+1}\)

\(\Leftrightarrow B-1=\dfrac{1011+1-1012-1}{1012+1}\)

\(\Leftrightarrow B-1=\dfrac{-1}{1012+1}\)

Ta có: \(1013+1>1012+1\)

\(\Leftrightarrow\dfrac{1}{1013+1}< \dfrac{1}{1012+1}\)

\(\Leftrightarrow\dfrac{-1}{1013+1}>\dfrac{-1}{1012+1}\)

\(\Leftrightarrow A-1>B-1\)

hay A>B

Vậy: A>B

so sánh phải ko bn

Yêu cầu đề bài là gì vậy bạn?

Số số hạng của dãy số trên là:

(1013 - 1) : 1 + 1 = 1013 (số)

Tổng của dãy số trên là:

(1013 + 1) x 1013 : 2 = 513591

Đ/:S...

tk mk nha

= (1011 - 1) + (1012 - 2) + (1013 - 3)

= 1010 + 1010 + 1010

= 1010 × 3 

= 3030

Ta có; \(B=1-\frac12+\frac13-\frac14+\cdots-\frac{1}{2022}+\frac{1}{2023}\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2023}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\ldots+\frac{1}{2023}-1-\frac12-\cdots-\frac{1}{1011}=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2023}\)

=C

=>B-C=0

Ta có: \(B=\frac12+\frac13-\frac14+\frac15-\frac16+\cdots-\frac{1}{2022}+\frac{1}{2023}\)

=>\(B=\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac{1}{2022}+\frac{1}{2023}-2\left(\frac14+\frac16+\cdots+\frac{1}{2022}\right)\)

\(=\frac12+\frac13+\frac14+\frac15+\cdots+\frac{1}{2022}+\frac{1}{2023}-\frac12-\frac13-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}+\frac{1}{2023}\)

=C

=>B-C=0

Ta có: \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2021\cdot2022}\)

\(=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\)

\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2022}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)

\(=1+\frac12+\frac13+\cdots+\frac{1}{2022}-1-\frac12-\cdots-\frac{1}{1011}\)

\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}\)

Ta có: \(B=1011+\frac{1010}{1012}+\frac{1009}{1013}+\cdots+\frac{2}{2020}+\frac{1}{2021}\)

\(=\left(\frac{1010}{1012}+1\right)+\left(\frac{1009}{1013}+1\right)+\cdots+\left(\frac{2}{2020}+1\right)+\left(\frac{1}{2021}+1\right)+1\)

\(=\frac{2022}{1012}+\frac{2022}{1013}+\cdots+\frac{2022}{2022}=2022\left(\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}\right)\)

=2022A

=>\(\frac{B}{A}=2022\) là số nguyên

Bài 1 :

\(6xy\cdot\sqrt{\frac{9x^2}{16y^2}}=6xy\cdot\frac{3x}{4y}=\frac{18x^2y}{4y}=\frac{9}{2}x^2\)

\(\sqrt{\frac{4+20a+25a^2}{b^4}}=\sqrt{\frac{\left(2+5a\right)^2}{\left(b^2\right)^2}}=\frac{2+5a}{b^2}\)

\(\left(m-n\right).\sqrt{\frac{m-n}{\left(m-n\right)^2}}=\sqrt{\left(m-n\right)^2}\cdot\sqrt{\frac{1}{m-n}}=\sqrt{\frac{\left(m-n\right)^2}{m-n}}=\sqrt{m-n}\)

Bài 2 : 

1. \(\left(2\sqrt{3}-\sqrt{12}\right):5\sqrt{3}=\left(2\sqrt{3}-2\sqrt{3}\right):5\sqrt{3}=0:5\sqrt{3}=0\)

2. \(\sqrt{\frac{317^2-302^2}{1013^2-1012^2}}=\frac{\sqrt{\left(317+302\right)\left(317-302\right)}}{\sqrt{\left(1013+1012\right)\left(1013-1012\right)}}=\frac{\sqrt{619}\cdot\sqrt{15}}{\sqrt{2025}}=\sqrt{\frac{619}{135}}\)(check lại)

3. \(\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{75}\)

\(=\sqrt{27}\left(1-\sqrt{3}\right):15\sqrt{3}\)

\(=3\sqrt{3}\left(1-\sqrt{3}\right):15\sqrt{3}\)

\(=\frac{1-\sqrt{3}}{5}\)

4.\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)

\(=\left(\frac{5}{\sqrt{5}}+\frac{\sqrt{20}}{2}-\frac{\frac{5}{4}\cdot2}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)

\(=\left(\sqrt{5}+\frac{2\sqrt{5}}{2}-\frac{\frac{5}{2}}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)

\(=\left(\sqrt{5}+\sqrt{5}+\frac{\sqrt{5}}{2}+\sqrt{5}\right):2\sqrt{5}\)

\(=\frac{7}{2}\sqrt{5}:2\sqrt{5}\)

\(=\frac{7}{4}\)