Tính nhanh
2004 x 2007 + 6 phần 2005 x 2005 + 2009
Tính nhanh
2004* 2007+ 6 / 2005* 2005 + 2009
\(=\dfrac{\left(2005-1\right)\left(2005+2\right)+6}{2005^2+2005+4}\)
\(=\dfrac{2005^2+2005-2+6}{2005^2+2005+4}=1\)
2004 x 2007 +6/ 2005 x 2005 + 2009
2004 x 2007 + 6 / 2005 x 2005 - 2009
2004x2007+6/2005x2005-2009=4020025
\(\frac{2004\times2007+6}{2005\times2005-2009}\)
\(=\frac{4022028+6}{4020025-2009}\)
\(=\frac{4022034}{4018016}.\)
Tính nhanh 2004*2007+6/2005*2005+2009
Tham khảo
Ta có:
2004×2007+6
=(2005-1)×2007+6
=2005×2007-2007+6
=2005×(2005+2)-2007+6
=2005×2005+2005×2-2007+6
=2005×2005+4010-2007+6
=2005×2005+9
Vậy 2004×2007+6/2005×2005+2009=1
\(=4022028+6+2009\)
\(=4022034+2099\)
\(=4024043\)
\(\dfrac{2004\text{×}2007+6}{2005\text{×}2005+2009}\)
\(=\dfrac{\left(2005-1\right)\text{×}2007+6}{2005\text{×}2005+2009}\)
\(=\dfrac{2005\text{×}2007-2007+6}{2005\text{×}2005+2009}\)
\(=\dfrac{2005\text{×}\left(2005+2\right)-2007+6}{2005\text{×}2005+2009}\)
\(=\dfrac{2005\text{×}2005+2005\text{×}2-2007+6}{2005\text{×}2005+2009}\)
\(=\dfrac{2005\text{×}2005+2009}{2005\text{×}2005+2009}=1\)
`#PhuonggYaa.`
tính giá trị của biểu thức 2007/2009×2002/2005×2009/2006×2005/2007×2006/2002
`2007/2009×2002/2005×2009/2006×2005/2007×2006/2002`
`=(2007xx2002xx2009xx2005xx2006)/(2009xx2005xx2006xx2007xx2002)`
`=(2007xx2002xx2009xx2005xx2006)/(2007xx2002xx2009xx2005xx2006)`
`=1`
\(\dfrac{2007}{2009}.\dfrac{2002}{2005}.\dfrac{2009}{2006}.\dfrac{2005}{2007}.\dfrac{2006}{2002}\\ =\left(\dfrac{2007}{2009}.\dfrac{2009}{2006}\right).\left(\dfrac{2006}{2002}.\dfrac{2002}{2005}\right).\dfrac{2005}{2007}\\ =\dfrac{2007}{2006}.\dfrac{2006}{2005}.\dfrac{2005}{2007}=1\)
\(\dfrac{2007}{2009}\text{×}\dfrac{2002}{2005}\text{×}\dfrac{2009}{2006}\text{×}\dfrac{2005}{2007}\text{×}\dfrac{2006}{2002}\\ =\dfrac{2007\text{×}2002\text{×}2009\text{×}2005\text{×}2006}{2009\text{×}2005\text{×}2006\text{×}2007\text{×}2002}=1\)
2004 nhân 2007 + 6/2005 nhân 2005 + 2009
Tính:(1+1/2005) x (1+ 1/2006) x (1+ 1/2007) x (1+ 1/2008) x (1 + 1/2009)
Giải phương trình:
(x+1)/(2010)+(x+2)/(2009)+(x+3)/(2008)=(x+4)/(2007)+(x+5)/(2006)+(x+6)/(2005)
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}=\frac{x+4}{2007}+\frac{x+5}{2006}+\frac{x+6}{2005}\)
<=> \(\frac{x+1}{2010}+1+\frac{x+2}{2009}+1+\frac{x+3}{2008}+1=\frac{x+4}{2007}+1+\frac{x+5}{2006}+1+\frac{x+6}{2005}+1\)
<=> \(\frac{x+2011}{2010}+\frac{x+2011}{2009}+\frac{x+2011}{2008}-\frac{x+2011}{2007}-\frac{x+2011}{2006}-\frac{x+2011}{2005}\) =0
<=> (x+2011).(\(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}\) )=0
<=> x+2011=0
<=> x=-2011
Vậy pt có nghiệm là x=-2011
Tính bằng cách thuật tiện :
2004 x 2007 + 6 / 2005 x 2005 +2009
Làm ra hộ mình nhé! Thanh you