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\(=\frac{\frac{5}{11.2}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{1}{11}+\frac{3}{2}}=\frac{5}{\frac{2}{4}}=\frac{5}{\frac{1}{2}}\)
a.\(\frac{1}{6}.6^x+6^x.36=6^{15}\left(1+6^3\right)\)
\(6^x.\frac{217}{6}=6^{15}.217\)
\(6^x=6^{16}\)
\(x=16\)
minh lam bai nay roi nhung minh ko co nha nen minh ko nho
Ta có :
(x+1)/2009 + (x+2)/2008 = (x+3)/2007 + (x+4)/2006
<=> (x+1)/2009 + 1 + (x+2)/2008 + 1 = (x+3)/2007 +1 + (x+4)/2006 + 1
<=> (x+2010)/2009 + (x+2010)/2008 = (x+2010)/2007 + (x+2010)/2006
<=> (x + 2010).[ 1/2009 + 1/2008 - 1/2007 - 1/2006 ] = 0
<=> x = -2010
| x | 7 | 9 | |||
| x2 | 49 | 81 | |||
| x2-49 | - | 0 | + | + | + |
| x2-81 | - | - | - | 0 | + |
| A | + | 0 | - | 0 | + |
dựa vào bảng ta có khi 7<x<9 thì A<0 vậy 7<x<9
b, ta có : \(\frac{2015}{1}\)+\(\frac{2014}{2}\)+\(\frac{2013}{3}\)+......+\(\frac{1}{2015}\)
=1+1+1+1......+1+\(\frac{2014}{2}\)+\(\frac{2013}{3}\)+.......+\(\frac{1}{2015}\)
(2015 số 1)
=1+(1+\(\frac{2014}{2}\))+(1+\(\frac{2013}{3}\))+........+(1+\(\frac{1}{2015}\))
=\(\frac{2016}{2016}\)+\(\frac{2016}{2}\)+\(\frac{2016}{3}\)+.........+\(\frac{2016}{2015}\)
=2016(\(\frac{1}{2016}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.........+\(\frac{1}{2015}\))
a: \(=\dfrac{17}{4}-\dfrac{37}{100}+\dfrac{1}{8}-\dfrac{32}{25}-\dfrac{5}{2}+\dfrac{7}{2}\)
\(=\dfrac{35}{8}+\dfrac{8}{8}-\dfrac{37}{100}-\dfrac{128}{100}\)
\(=\dfrac{43}{8}-\dfrac{165}{100}=\dfrac{149}{40}\)
b: \(=\left(\dfrac{22\cdot26+3\cdot10-65}{130}\right):\left(\dfrac{4\cdot22-2\cdot26+3\cdot143}{286}\right)\)
\(=\dfrac{537}{130}\cdot\dfrac{286}{465}=\dfrac{1969}{775}\)