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Giải bài khó nhất =)
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
Do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\ne0\) nên \(x+2004=0\Leftrightarrow x=-2004\)
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
nên x + 1 = 0 => x = -1
Vậy x = -1
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}=\frac{2004+x}{2002}+\frac{2004+x}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}-\frac{2004+x}{2002}-\frac{2004+x}{2003}=0\)
\(\left(2004+x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)
nên 2004 + x = 0 => x = -2004
Vậy x = -2004
=))
Có \(\frac{x+4}{2000}\) + \(\frac{x+3}{2001}\) = \(\frac{x+2}{2002}\) + \(\frac{x+1}{2003}\)
( \(\frac{x+4}{2000}\) + 1 ) + ( \(\frac{x+3}{2001}\) + 1 ) = ( \(\frac{x+2}{2002}\) + 1 ) + ( \(\frac{x+1}{2003}\) + 1 )
( \(\frac{x+4}{2000}\) + \(\frac{2000}{2000}\) ) + ( \(\frac{x+3}{2001}\) + \(\frac{2001}{2001}\) ) = ( \(\frac{x+2}{2002}\) + \(\frac{2002}{2002}\) ) + ( \(\frac{x+1}{2003}\) + \(\frac{2003}{2003}\) )
\(\frac{x+4+2000}{2000}\) + \(\frac{x+3+2001}{2001}\) = \(\frac{x+2+2002}{2002}\) + \(\frac{x+1+2003}{2003}\)
\(\frac{x+2004}{2000}\) + \(\frac{x+2004}{2001}\) = \(\frac{x+2004}{2002}\) + \(\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}\) + \(\frac{x+2004}{2001}\) - \(\frac{x+2004}{2002}\) - \(\frac{x+2004}{2003}\) = 0
( x + 2004 ) + ( \(\frac{1}{2000}\) + \(\frac{1}{2001}\) + \(\frac{1}{2002}\) + \(\frac{1}{2003}\) ) = 0
Mà \(\frac{1}{2000}\) + \(\frac{1}{2001}\) + \(\frac{1}{2002}\) + \(\frac{1}{2003}\) \(\ne\) 0
\(\Rightarrow\) x + 2004 = 0
\(\Rightarrow\) x = -2004
Vậy x = - 2014
\(\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}=\frac{x-4}{2001}\)
\(\Rightarrow\frac{x-1}{2004}+\frac{x-2}{2003}-\frac{x-3}{2002}-\frac{x-4}{2001}=0\)
\(\Rightarrow\frac{x-1}{2004}-1+\frac{x-2}{2003}-1-\frac{x-3}{2002}+1-\frac{x-4}{2001}+1=0\)
\(\Rightarrow\left(\frac{x-1}{2004}-1\right)+\left(\frac{x-2}{2003}-1\right)-\left(\frac{x-3}{2002}-1\right)-\left(\frac{x-4}{2001}-1\right)=0\)
\(\Rightarrow\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
\(\Rightarrow\left(x-2005\right).\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Vì \(\frac{1}{2004}< \frac{1}{2002};\frac{1}{2003}< \frac{1}{2001}\)\(\Rightarrow\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\ne0\)
\(\Rightarrow x-2005=0\)
\(\Rightarrow x=2005\)
Vậy x = 2005
x-1/65-1-x-3/63-1+x-5/61-1+x-7/59-1 x-66/65-x-66/63+x-66/61+x-66/59 =0 suy ra (x-66).(1/65-1/63+1/61+1/59)=0 vi 1/65-1/63+1/61+1/59khong thuoc 0 nen x-66+66=0 suy ra x =132
a-b=2.(a+b) tương đương a-b =2a + 2b tương đương -3b=a
a-b=a.b suy ra -3b-b=-3b.b tương đương -4b=-3b.b tương đương b=4/3 suy ra a=-4
với a=-4 ; b=4/3 thì a-b = 2.(a+b)= a.b
\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)-\left(\frac{x+2}{2002}+1\right)-\left(\frac{x+1}{2003}+1\right)=0\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2000}>\frac{1}{2001}>\frac{1}{2002}>\frac{1}{2003}\)
=>\(x+2004=0\)
=>\(x=-2004\)
a, Ta có \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{1+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
=> \(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right).\left(\frac{1}{13}+\frac{1}{14}\right)\)
=> \(\left(x+1\right).\frac{181}{660}=\left(x+1\right).\frac{27}{182}\)
=> \(x.\frac{181}{660}+\frac{181}{660}=x.\frac{27}{182}+\frac{27}{182}\)
=> \(x\frac{181}{660}-x\frac{27}{182}=\frac{27}{182}-\frac{181}{660}\)
=> \(x\frac{7561}{60060}=-\frac{7561}{60060}\)
=> x = -1
Mk chỉ biết cách làm này thôi mặc dù có cách khác nữa ( bạn thông càm )
Tích cho mk nhé