\(\dfrac{2010\cdot125+1000}{126\cdot2010-1010}\)
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\(\frac{2010\cdot2011+1000}{2012\cdot2010-1010}\)
= \(\frac{2010\cdot2011+1000}{\left(2011+1\right)\cdot2010-1010}\)
= \(\frac{2010\cdot2011+1000}{2011\cdot2010+2010-1010}\)
= \(\frac{2010\cdot2011+1000}{2011\cdot2010+1000}\)
= 1
\(\frac{2010.2011+1000}{2012.2010-1010}\)
\(=\frac{2010.2011+2010-1010}{2012.2010-1010}\)
\(=\frac{2010.\left(2011+1\right)-1010}{2012.2010-1010}\)
\(=\frac{2010.2012-1010}{2012.2010-1010}\)
\(=1\)
\(\frac{2010.125+1000}{126.2010-1010}=\frac{10\left(201.125+100\right)}{10\left(201.126-101\right)}=\frac{201.125-101+201}{2011.126-101}\)
\(=\frac{201.126-101}{201.126-101}=1\)
=2010 x 125 + 2010x 126 + 1000+1010
=2010 x 125 + 2010 x 126 + 2010
=2010 x ( 125 + 126 +1 )
=2010 x 252
=506520
\(a,=15\left(\dfrac{2121}{4343}+\dfrac{222222}{434343}\right)=15\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15\cdot1=15\)
Ta có :
Tử số :
( 2005 + 1 ) . 125 + 1000
2005 . 125 + 125 + 1000
2005 . 125 +1125
Mẫu Số :
( 125 + 1 ) . 2005 - 888
125 . 2005 + 2005 - 888
125 . 2005 + 1117
Ta có phân số : \(\dfrac{2005\cdot125+1125}{125\cdot2005+1117}\) = \(\dfrac{1125}{1117}\)
\(\frac{\left(2005+1\right).125+1000}{\left(125+1\right).2005-888}\)
= \(\frac{2005.125+125+1000}{125.2005+2005-888}\)
= \(\frac{2005.125+1125}{125.2005+1117}\)
= \(\frac{250625+1125}{250625+1117}\)
= \(\frac{125875}{125871}\)
A=1⋅2⋅3⋅...⋅2010(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+...+\(\dfrac{1}{2010}\))
= 1⋅2⋅3⋅...⋅2010[(1+\(\dfrac{1}{2010}\))+(\(\dfrac{1}{2}\)+\(\dfrac{1}{2009}\))+(\(\dfrac{1}{3}\)+\(\dfrac{1}{2008}\))+...+(\(\dfrac{1}{1005}\)+\(\dfrac{1}{1006}\))]
= 1⋅2⋅3⋅...⋅2010(\(\dfrac{2011}{2010}\)+\(\dfrac{2011}{2009\cdot2}\)+\(\dfrac{2011}{2008\cdot3}\)++...+\(\dfrac{2011}{1006\cdot1005}\))
= 2011*(\(\dfrac{2010!}{2010}\)+\(\dfrac{2010!}{2009\cdot2}\)+\(\dfrac{2010!}{2008\cdot3}\)++...+\(\dfrac{2010!}{1006\cdot1005}\))
=> A⋮2011 (dpcm)
Bạn nguyen quang huy sai rồi!!!
Vì 1000/2009>1000/2009+2010 (1)
1010/2010>1010/2009+2010 (2)
Ta cộng theo vế (1) và (2) với nhau nên ta được:
1000/2009+1010/2010>1000/2009+2010 +1010/2009+2010
=>1000/2009+1010/2010>1000+1010/2009+2010
Vậy A<B
Chắc chắn 100% luôn, không sai đâu!!!!!!!
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{21^2\cdot14\cdot125}{35^2\cdot125}\)
`=`\(\dfrac{3^2\cdot7^2\cdot2\cdot7\cdot5^2}{5^2\cdot7^2\cdot5^2}\)
`=`\(\dfrac{3^2\cdot2\cdot7\cdot5^2\cdot7^2}{5^2\cdot5^2\cdot7^2}\)
`=`\(\dfrac{3^2\cdot2\cdot7}{5^2}=\dfrac{126}{25}\)
Lời giải:
$\frac{2010\times 125+1000}{126\times 2010-1010}=\frac{2010\times (126-1)+1000}{126\times 2010-1010}$
$=\frac{2010\times 126-(2010-1000)}{126\times 2010-1010}=\frac{2010\times 126-1010}{126\times 2010-1010}=1$
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