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quá đơn giản
n = 1
0 x {2008 x 2009 - 2006 x 2007} = 0
vậy 2009 x 1 - 2009 =0
{2009 x n - 2009} : {2008 x 2009 - 2006 x 2007} = 0 => {2009 x n - 2009} = 0 => n= 1
2009 + 2008 + 2007 + ..... + (x + 1) + x = 2009
x + (x + 1) + (x + 2) + .......... + 2008 + 2009 = 2009
Áp dụng công thức tính dãy số ta có :
\(\frac{\left[\left(2009-x\right):1+1\right].\left(2009+x\right)}{2}=2009\)
\(\frac{\left[2009-x+1\right]\left(2009+x\right)}{2}=2009\)
\(\left[2008-x\right]\left(2009+x\right)=4018\)
\(2008\left(2009+x\right)-x\left(2009+x\right)=4018\)
\(2008.2009+2008x-\left(2009x+x^2\right)=4018\)
2008.2009 + 2008x - 2009x - x2 = 4018
2008.2009 - x - x2 = 4018
2008.2009 - x(x + 1) = 4018
x(x + 1) = 4034072 - 4018
x(x + 1) = 4030054
Còn lại cậu dò tìm số x là được !!!
a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)
\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)
\(=\dfrac{2}{5}+\dfrac{4}{2}\)
\(=\dfrac{2}{5}+2\)
\(=\dfrac{2}{5}+\dfrac{10}{5}\)
\(=\dfrac{12}{5}\)
b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)
\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)
\(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)
\(=1-\dfrac{2}{2008}\)
\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)
\(=\dfrac{2006}{2008}\)
\(=\dfrac{1003}{1004}\)
a: =2/5+4/2
=2/5+2
=12/5
b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)
\(=\left(\frac{2007}{2008}+\frac{1}{2008}\right)+\left(\frac{2007}{2008}.\frac{2008}{2009}+\frac{1}{2009}\right)\)
\(=1+\frac{2008}{2009}=\frac{4017}{2009}\)
\(\frac{2007}{2008}+\frac{1}{2009}+\frac{2007}{2008}:\frac{2009}{2008}+\frac{1}{2008}\)
\(=\frac{2007}{2008}+\frac{1}{2009}+\frac{2007}{2009}+\frac{1}{2008}\)
\(=\left(\frac{2007}{2008}+\frac{1}{2008}\right)+\left(\frac{2007}{2009}+\frac{1}{2009}\right)\)
\(=1+\frac{2008}{2009}\)
\(=\frac{4017}{2009}\)
\(\frac{2009x2008-1}{2007x2009+2008}=\frac{2009x2007+2009-1}{2009x2007+2008}=1.\)
vậy biểu thức trên =1
......................... hihi !