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![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: \(\frac{-4}{9}=\frac{-8}{18}>\frac{-8}{13}\Rightarrow\frac{-4}{9}>\frac{-8}{13}\)
b,Ta có: \(\frac{-2005}{2006}>-1\)
\(\frac{-2007}{2004}=-1-\frac{3}{2004}-1>-\frac{2007}{2004}\)
Vậy -2005/2006>-2007/2004
![](https://rs.olm.vn/images/avt/0.png?1311)
a.\(\frac{13}{17}\)=1-\(\frac{4}{17}\); \(\frac{46}{50}\)=1-\(\frac{4}{50}\)
Vì \(\frac{4}{17}\)>\(\frac{4}{50}\)=> 1-\(\frac{4}{17}\)<1-\(\frac{4}{50}\)
Vậy\(\frac{13}{17}\)<\(\frac{46}{50}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Quy đồng : \(\frac{-33}{134}=\frac{-33.203}{134.203}=\frac{-6699}{134.203}\)
\(\frac{-51}{203}=\frac{-52.134}{203.134}=\frac{-6834}{134.203}\)
So sánh tử ta được -6699>-6834
Nên: \(\frac{-6699}{134.203}>\frac{-6834}{134.203}\) ( Với hai số cùng mẫu, tử càng lớn thì phân số càng lớn)
Vậy \(\frac{33}{-134}>\frac{-51}{203}\)
Ta so sánh \(\frac{33}{134}\)với\(\frac{51}{203}\)
Ta có:
\(\frac{33}{134}< \frac{33}{132}=\frac{1}{4}\)
\(\frac{51}{203}>\frac{51}{204}=\frac{1}{4}\)
\(\Rightarrow\frac{33}{134}< \frac{1}{4}< \frac{51}{203}\)
\(\Rightarrow\frac{33}{-134}>\frac{1}{4}>\frac{-51}{203}\)
\(.\Rightarrow\frac{33}{-134}>\frac{-51}{203}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\frac{33}{-37}=\frac{-33}{37}\)
Mà \(\frac{-33}{37}>\frac{-34}{37}>\frac{-34}{35}\)=>\(\frac{-33}{37}>\frac{-34}{35}\)
Vậy \(\frac{33}{-37}>\frac{-34}{35}\)
Vì \(\frac{2019}{2020}=1-\frac{2019}{2020}=\frac{1}{2020}\)
\(\frac{2006}{2007}=1-\frac{2006}{2007}=\frac{1}{2007}\)
Vì \(\frac{1}{2020}< \frac{1}{2007}=>\frac{2019}{2020}>\frac{2006}{2007}\)
Vậy\(\frac{2019}{2020}>\frac{2006}{2007}\)
Chiều nay mình giải tiếp cho.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(\Rightarrow P=\frac{1\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2002}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)
\(\Rightarrow P=\frac{-7}{15}\)
Vậy \(P=\frac{-7}{15}\)
Bài 2:
Ta có: \(S=23+43+63+...+203\)
\(\Rightarrow S=13+10+20+23+...+103+100\)
\(\Rightarrow S=\left(13+23+...+103\right)+\left(10+20+...+100\right)\)
\(\Rightarrow S=3025+450\)
\(\Rightarrow S=3475\)
Vậy S = 3475
1. \(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
=> P =\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
=> P = \(\frac{1}{5}-\frac{2}{3}\)
P = \(\frac{3}{15}-\frac{10}{15}\)
=> P =\(\frac{-7}{15}\)
2. ta có:
S = 23 + 43 + 63 +...+ 203
=> S = 13 + 10 + 23 + 20 +...+ 103 + 100
=> S = ( 13 + 23+...+ 103 ) + ( 10 + 20 +...+ 100 )
=> S = 3025 + 550
=> S = 3575
Vậy S = 3575
\(\frac{4}{-9}=-\frac{4}{9}\)
\(\frac{8}{-13}=-\frac{8}{13}\)
MC=117
Quy đồng:
\(-\frac{4}{9}=-\frac{52}{117}\)
\(-\frac{8}{13}=-\frac{72}{117}\)
=>Vì -52>-72, nên \(-\frac{52}{117}>-\frac{72}{117}\)hay \(\frac{4}{-9}>\frac{8}{-13}\)