2011 x 2010 - 1/2009 x 2011 + 2010
K
Khách
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TH
2
22 tháng 2 2020
\(\frac{x-2010-2011}{2009}+\frac{x-2009-2011}{2010}+\frac{x-2009-2010}{2011}=3\)
\(\Leftrightarrow\left(\frac{x-2010-2011}{2009}-1\right)+\left(\frac{x-2009-2011}{2010}-1\right)+\left(\frac{x-2009-2010}{2011}-1\right)=0\)
\(\Leftrightarrow\frac{x-6030}{2009}+\frac{x-6030}{2010}+\frac{x-6030}{2011}=0\)
\(\Leftrightarrow\left(x-6030\right)\left(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}\right)\)
\(\Leftrightarrow x-6030=0\)(vì \(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}>0\))
\(\Leftrightarrow x=6030\)
Vậy ................
BS
0
13 tháng 2 2019
=\(\dfrac{1}{2009.\left(\dfrac{1}{2009}+\dfrac{1}{2011}+\dfrac{1}{2010}\right)}+\dfrac{1}{2010.\left(\dfrac{1}{2010}+\dfrac{1}{2009}+\dfrac{1}{2011}\right)}+\dfrac{1}{2011.\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2010}\right)}\)\(=\dfrac{1}{2009}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2010}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)+\dfrac{1}{2011}:\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)\)
\(=\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right):\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2011}\right)=1\)
TN
0
NM
Giaỉ phương trình:
\( \sqrt{x-2009}-1/{x-2009}+ \sqrt{y-2010}-1/y-2010+ \sqrt{z-2011}-1/z-2011 =3/4\)
2
NP
8 tháng 4 2015
−1x−2009+y−2010−−−−−−−√−1y−2010+z−2011−−−−−−−√−1z−2011=34
Ta có
x−2009−−−−−−−√−1x−2009+y−2010−−−−−−−√−1y−2010+z−2011−−−−−−−√−1z−2011=34⇔(1x−2009−−−−−−−√−12)2+(1y−2010−−−−−−−√−12)2+(1z−2011−−−−−−−√−12)2=0
⇒x=2013,y=2014,z=2015
`(2011xx2020-1)/(2009xx2011+2010)`
`=((2009+1)xx2011-1)/(2009xx2011+2010)`
`=(2009xx2011+2011-1)/(2009xx2011+2010)`
`=(2009xx2011+2010)/(2009xx2011+2010)`
`=1`
\(\dfrac{2011.2010-1}{2009.2011+2010}\)
= \(\dfrac{2011.2009+2011-1}{2009.2011+2010}\)
= \(\dfrac{2011.2009+2010}{2009.2011+2010}\)
= 1