< ...hên sui nhe...:))

Không quy đồng hả sao khó quá z:

\(\frac{-7}{10^{2013}}\)+\(\frac{-15}{10^{2014}}\)  >   \(\frac{-15}{10^{2013}}\)+\(\frac{-7}{10^{2014}}\)

\(A=\frac{-7}{10^{2012}}+\frac{-15}{10^{2013}}=\frac{-15+8}{10^{2012}}+\frac{-15}{10^{2013}}=\frac{-15}{10^{2012}}+\frac{-15}{10^{2013}}+\frac{8}{10 ^{2012}}\);

\(B=\frac{-15}{10^{2012}}+\frac{-7}{10^{2013}}=\frac{-15}{10^{2012}}+\frac{-15+8}{10^{2013}}=\frac{-15}{10^{2012}}+\frac{-15}{10^{2013}}+\frac{8}{10^{2013}}\);

mà \(\frac{8}{10^{2012}}>\frac{8}{10^{2013}}\Rightarrow A>B\)

A= \(\frac{8-15}{10^{2014}}+\frac{-15}{10^{2015}}=\frac{8}{10^{2014}}-15.\left(\frac{1}{10^{2014}}+\frac{1}{10^{2015}}\right)\)

B=\(\frac{-15}{10^{2014}}+\frac{8-15}{10^{2015}}=\frac{8}{10^{2015}}-15.\left(\frac{1}{10^{2014}}+\frac{1}{10^{2015}}\right)\)

Nhận xét: \(\frac{8}{10^{2015}}

B < A

mới lập níc,k cho mình nhá!

\(A=\frac{8-15}{10^{2014}}+\frac{-15}{10^{2015}}=\frac{8}{10^{2014}}-15.\left(\frac{1}{10^{2014}}+\frac{1}{10^{2015}}\right)\)

\(B=\frac{-15}{10^{2014}}+\frac{8-15}{10^{2015}}=\frac{8}{10^{2015}}-15.\left(\frac{1}{10^{2014}}+\frac{1}{10^{2015}}\right)\)

Ta có nhận xét :

\(\frac{8}{10^{2015}}< \frac{8}{10^{2014}}\Rightarrow\frac{8}{10^{2015}}-15.\left(\frac{1}{10^{2014}}+\frac{1}{10^{2015}}\right)< \frac{8}{10^{2014}}-15.\left(\frac{1}{10^{2014}}+\frac{1}{10^{2015}}\right)\)

\(\Rightarrow B< A\) 

a) Ta có: \(12 = 2^2 . 3; 15 = 3.5\)

\(BCNN(12, 15) = 2^2.3.5 = 60\) nên chọn mẫu chung là 60.

\(\begin{array}{l}\frac{9}{{12}} = \frac{{9.5}}{{12.5}} = \frac{{45}}{{60}}\\\frac{7}{{15}} = \frac{{7.4}}{{15.4}} = \frac{{28}}{{60}}\end{array}\)

b) Ta có: \(10 = 2.5; 4 = 2^2; 14=2.7\)

\(BCNN(10, 4, 14) =2^2.5.7= 140\) nên chọn mẫu chung là 140.

\(\begin{array}{l}\frac{7}{{10}} = \frac{{7.14}}{{10.14}} = \frac{{98}}{{140}}\\\frac{3}{4} = \frac{{3.35}}{{4.35}} = \frac{{105}}{{140}}\\\frac{9}{{14}} = \frac{{9.10}}{{14.10}} = \frac{{90}}{{140}}\end{array}\)

Ta có

\(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)

\(B=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)

Vì \(\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\)

=>A>B

quy dong ca A va B ta dc :

\(A=\frac{-109}{10^{2014}}\)

\(B=\frac{-199}{10^{2014}}\)

\(\Rightarrow A>B\)

dễ thôi

ta có :A=-9/10^2013+-19/10^2014=-9/10^2013+-9/10^2014+-10/10^2014

          B=-9/10^2014+-19/10^2013=-9/10^2014+-9/10^2013+-10/10^2013

nhìn nhé :cả A và B đều có các số hạng :-9/10^2013 và-9/10^2014

mà -10/10^2014<-10/10^2013

=>A<B

b,Ta có 

\(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)

\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)

\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)

\(\Rightarrow P>Q\)

\(A=\frac{-10}{20}+\frac{-10}{30}+\frac{-10}{42}+\frac{-10}{56}+\frac{-10}{72}+\frac{-10}{90}+\frac{-10}{110}\)

\(=-10\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)\)

\(=-10\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)

\(=-10\left(\frac{1}{4}-\frac{1}{11}\right)\)

\(=\frac{-35}{22}\)