\(\frac{2011.4023+2012}{2012.4023-2011}=\frac{2011.4023+2011+1}{2012.4023-2012-1}=\frac{2011.4023+2011.1+1}{2012.4023-2012.1-1}\)

\(=>\frac{2012.4023+2012.1+1}{2012.4023-2012.1-1}=\frac{2012.\left(4023+1\right)+1}{2012.\left(4023-1\right)-1}\)

\(=\frac{4023+1+1}{4023-1-1}=\frac{4023+2}{4023-2}=\frac{4025}{4021}\)

Vì 4025 > 4021 ( tử số lớn hơn mẫu số ) nên suy ra : 4025/4021 >1

<br class="Apple-interchange-newline"><div id="inner-editor"></div>=>2012.4023+2012.1+12012.4023−2012.1−1 =2012.(4023+1)+12012.(4023−1)−1 

=4023+1+14023−1−1 =4023+24023−2 =40254021 

Vì 4025 > 4021 ( tử số lớn hơn mẫu số ) nên suy ra : 4025/4021 >1

\(\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right)......\left(1-\frac{1}{2011}\right)\)

\(=\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2010}{2011}\)

\(=\frac{6.7.8.9.....2010}{7.8.9.10.....2011}\)

\(=\frac{6}{2011}\)

\(VP=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)

\(=1-1+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{40024}{2012}-1\right)+2012\)

\(=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}+\frac{2012}{1}\)

\(=2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)

\(\Rightarrow2012=503.x\Rightarrow x=\frac{2012}{503}=4\)

a) Ta có :

\(A=\frac{10^{2010}+1}{10^{2011}+1}\)

\(\Rightarrow10A=\frac{10^{2011}+10}{10^{2011}+1}=\frac{\left(10^{2011}+1\right)+9}{10^{2011}+1}=1+\frac{9}{10^{2011}+1}\)

\(B=\frac{10^{2011}+1}{10^{2012}+1}\)

\(\Rightarrow10B=\frac{10^{2012}+10}{10^{2012}+1}=\frac{\left(10^{2012}+1\right)+9}{10^{2012}+1}=1+\frac{9}{10^{2012}+1}\)

Vì \(\frac{9}{10^{2011}+1}>\frac{9}{10^{2012}+1}\)nên \(10A>10B\)

\(\Rightarrow A>B\)

Vậy : \(A>B\)

b) Ta có :

\(\left(\frac{-1}{2}\right)^{11}=\frac{-1^{11}}{2^{11}}=\frac{-1}{2^{11}}\)

\(\left(\frac{-1}{2}\right)^{13}=\frac{-1^{13}}{2^{13}}=\frac{-1}{2^{13}}\)

Vì \(\frac{-1}{2^{11}}>\frac{-1}{2^{13}}\)nên \(\left(\frac{-1}{2}\right)^{11}>\left(\frac{-1}{2}\right)^{13}\)

Vậy : \(\left(\frac{-1}{2}\right)^{11}>\left(\frac{-1}{2}\right)^{13}\)

\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+1+9}{10^{2012}+1+9}\)

\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10^{2011}+10}{10^{2012}+10}\)

\(B=\frac{10^{2011}+1}{10^{2012}+1}< \frac{10\cdot\left(10^{2010}+1\right)}{10\cdot\left(10^{2011}+1\right)}=\frac{10^{2010}+1}{10^{2011}+1}=A\)

Vậy : B < A

\(A=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{12}\right)...\left(1-\frac{1}{2007}\right)\left(1-\frac{1}{2008}\right)\)

     \(=\frac{9}{10}.\frac{10}{11}.\frac{11}{12}.....\frac{2006}{2007}.\frac{2007}{2008}\)

     \(=\frac{9.10.11.....2006.2007}{10.11.12.....2007.2008}\)

     \(=\frac{9}{2008}\)

\(Ta\) \(có:\)

\(A=\frac{9}{2008}\)

\(B=\frac{1}{2000}\)

\(\frac{9}{2008}=\frac{9.250}{2008.250}=\frac{2250}{502000}\)

\(\frac{1}{2000}=\frac{1.251}{2000.251}=\frac{251}{502000}\)

Vì \(\frac{2250}{502000}>\frac{251}{502000}\Rightarrow A>B\)

\(A=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{12}\right)...\left(1-\frac{1}{2007}\right)\left(1-\frac{1}{2008}\right)\)

\(A=\frac{9}{10}.\frac{10}{11}.\frac{11}{12}....\frac{2006}{2007}.\frac{2007}{2008}\)

\(A=\frac{9.10.11....2006.2007}{10.11.12...2007.2008}\)

\(A=\frac{9}{2008}\)

Vì \(\frac{9}{2008}

\(D=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right):\left(\frac{2011}{1}+\frac{2010}{2}+...+\frac{1}{2011}\right)\)

\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)

\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)

\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)

\(\Rightarrow D\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)

\(\Rightarrow D=\frac{1}{2012}\)