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b, \(\frac{x+1}{2009}+\frac{x+2}{2009}=\frac{x+10}{2000}+\frac{x+11}{1999}\)
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)\)
\(\Rightarrow\frac{x+1+2009}{2009}+\frac{x+2+2008}{2008}=\frac{x+10+2000}{2000}+\frac{x+11+1999}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2000}+\frac{x+2010}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2000}-\frac{x+2010}{1999}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\right)=0\)
Mà \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\ne0\)
=> x + 2010 = 0 => x = -2010
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\(\frac{2009x2008-1}{2007x2009+2008}=\frac{2009x2007+2009-1}{2009x2007+2008}=1.\)
vậy biểu thức trên =1
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Điều kiện: 2009.x\(\ge\)0
\(\Rightarrow\)x\(\ge\)0
\(\Rightarrow\)/x+1/, /x+2/, .... , /x+2008/\(\ge\)0
\(\Rightarrow\)/x+1/+/x+2/+...+/x+2008/=2009.x\(\Leftrightarrow\)2008x+1+2+...+2008=2009x
\(\Rightarrow\)x=2017036
Vậy \(x=2017036\)
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a.(x-1)(x-5)=-18
<=>x2-5x-x+5=-18
<=>x2-5x-x+5+18=0
<=>x2-6x+23=0
Denta =(-6)2-4.1.23=-56 PTVN
b.5(x-2)(x+3)=15
<=>5(x2+3x-2x-6)=15
<=> 5x2 +15x-10x-30=15
<=> 5x2+15x-10x-30-15
<=> 5x2 +5x -45=0
Denta = 52-4.5.(-45)=525>0
Phương trình có 2 nghiệm phân biệt
x1= -5+√525/10=....
x2=-5-√525/10=....