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2010.2014
=(2012-2)(2012+2)
=2012.2012-4
=2012^2-4
=> 2012^2>2010.2014.
EZ mà bn.
Có : \(2010.2014=\left(2012-2\right)\left(2012+2\right)\)
\(=2012^2-4< 2012^2\)
Vậy ................
\(A=\frac{199}{200}+\frac{200}{201}+\frac{201}{202}< \frac{199}{200+201+202}+\frac{200}{200+201+202}+\frac{201}{200+201+202}\)
A \(< \frac{199+200+201}{200+201+202}=B\)
\(A< B\)
Ta có: \(A=\frac{199}{200}+\frac{200}{201}+\frac{201}{202}< \frac{199}{200+201+202}+\frac{200}{200+201+202}+\frac{201}{200+201+202}< \)
\(< \frac{199+200+201}{200+201+202}\)
Vậy A < B
ỦNG HỘ TỚ NHA
\(\frac{199}{200}>\frac{199}{200+201+202}\)
\(\frac{200}{201}>\frac{200}{200+201+202}\)
\(\frac{201}{202}>\frac{201}{200+201+202}\)
=>\(A>B\)
Do \(\frac{199}{200}\)> \(\frac{199}{200+201+202}\), \(\frac{200}{201}\)>\(\frac{200}{200+201+202}\),\(\frac{201}{202}\)>\(\frac{201}{200+201+202}\)nên A>B
2) \(\frac{2010+2011}{2011+2012}=\frac{2010}{2011+2012}+\frac{2011}{2011+2012}\)
\(\frac{2010}{2011}>\frac{2010}{2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)
\(\Rightarrow A>B\)
- Bài 3:
a) \(1\frac{13}{15}.\left(0,5\right)^2.3+\left(\frac{8}{15}-1\frac{19}{60}\right):1\frac{23}{24}\)
\(=\frac{28}{15}.\frac{1}{4}.3+-\frac{47}{60}:\frac{47}{24}\)
\(=\frac{7}{5}-\frac{2}{5}\)
\(=1\)
b)\(\frac{\left(\frac{11^2}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+3\frac{1}{6}}\)
\(=\frac{\left(\frac{121}{200}+0,415\right):0,01}{\frac{1}{12}-37,25+\frac{19}{6}}\)
\(=\frac{\frac{51}{50}:\frac{1}{100}}{-34}\)
\(=\frac{102}{-34}\)
\(=-3\)
Ta có:
E=35.53 - 18 và F= 35 + 53.34
= (34 + 1) . 53 - 18
= 34 . 53 + 53 - 18
= 34 . 53 + 35
=> E=F hay 35.53 - 18 = 35 + 53 . 34
Vậy 35.53 - 18 = 35 + 53 .34
Ta có:
a) \(A=2000\cdot2009=2000\cdot\left(2005+4\right)=2000\cdot2005+2000\cdot4\)
\(B=2004\cdot2005=\left(2000+4\right)\cdot2005=2000\cdot2005+2005\cdot4\)
Do \(2000< 2005\)
\(\Rightarrow2000\cdot4< 2005\cdot4\)
\(\Rightarrow2000\cdot2005+2000\cdot4< 2000\cdot2005+2005\cdot4\)
Vậy A < B
b) \(A=3004^2=\left(3000+4\right)\cdot3004=3000\cdot3004+3004\cdot4\)
\(B=3000\cdot3008=3000\cdot\left(3004+4\right)=3000\cdot3004+3000\cdot4\)
Do \(3004>3000\)
\(\Rightarrow3004\cdot4>3000\cdot4\)
\(\Rightarrow3000\cdot3004+3004\cdot4>3000\cdot3004+3000\cdot4\)
Vậy A > B
b,Ta có
\(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
\(\Rightarrow P>Q\)
\(A=\frac{-10}{20}+\frac{-10}{30}+\frac{-10}{42}+\frac{-10}{56}+\frac{-10}{72}+\frac{-10}{90}+\frac{-10}{110}\)
\(=-10\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)\)
\(=-10\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=-10\left(\frac{1}{4}-\frac{1}{11}\right)\)
\(=\frac{-35}{22}\)
a)
A=2012.2012
=(2010+2).2012
=2010.2012+4024
B=2010.2014
=2010.(2012+2)
=2010.2012+4020
Vì 2010.2012+4024>2010.2012+4020 Nên A>B
b) C=199.201
=199.(200+1)
=199.200+199
D=200.200
=(199+1).200
=199.200+200
Vì 199.200+199<199.200+200 nên C<D
**** chị nha emmmmmmmmmmm