Tính tử: \(51.125-51.42-17.150\)

\(=51.125-51.42-17.3.50\)

\(=51.125-51.42-51.50\)

\(=51.\left(125-42-50\right)=51.33\)

Tính mẫu: 

Số các số hạng: \(\left(99-3\right):3+1=33\)

Mẫu = \(33.\left(\dfrac{99+3}{2}\right)=33.51\)

\(B=1\)

Ta có: \(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=0\)

\(\dfrac{\left[\left(x+1\right)+\left(x+99\right)\right].50}{2}=0\)

\(\left(x+50\right).50=0\)

\(x+50=0\)

\(x=-50\)

chữ xấu (thông cảm) ;-;

Ta có \(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2021}\left(1\right)\)

\(\Rightarrow\dfrac{3}{2}A=\dfrac{3}{4}+\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\left(\dfrac{3}{2}\right)^4+...+\left(\dfrac{3}{2}\right)^{2013}\left(2\right)\)

Lấy (2) - (1) ta được:

\(\dfrac{3}{2}A-A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}\)

\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}+\dfrac{1}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}+\dfrac{1}{2}\)

Vậy \(B-A=\dfrac{3^{2013}}{2^{2014}}-\dfrac{3^{2013}}{2^{2012}}+\dfrac{5}{2}\)

\(1,25:\dfrac{15}{20}+\left(25\%-\dfrac{5}{6}\right):4\dfrac{2}{3}\)

\(=\dfrac{5}{4}.\dfrac{4}{3}+\left(\dfrac{1}{4}-\dfrac{5}{6}\right).\dfrac{3}{14}\)

\(=\dfrac{5}{3}+\dfrac{7}{12}.\dfrac{3}{14}\)

\(=\dfrac{5}{3}+-\dfrac{1}{8}\)

\(=\dfrac{37}{24}\)

1. need

2. won't be

3. will buy

4. will write

5. will go

\(\dfrac{13}{x-15}\) là số nguyên khi \(x-15\) là ước của 13

\(x-15\in\left\{\pm1;\pm13\right\}\)

\(\Rightarrow x\in\left\{16;14;26;2\right\}\)

Ta có:

 \(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)

\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)

\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)

Do đó: \(\dfrac{2008}{2009}+\dfrac{2009}{2010}+\dfrac{2010}{2011}>\dfrac{2008+2009+2010}{2009+2010+2011}\)

\(\Rightarrow A>B\)