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\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
Ta có:(1+1999/2)+(1+1998/3)+...(2/1999)(có 1998 tổng<=>1998 số 1)+(2000 - 1998)+400
= 2001/2+2001/3+...+2001/1999+402
=2001.(1/2+1/3+...+1/1999)+402(1)
Thay (1) vào biểu thức trên và tính(tự tính nha!,tk cho mk!!!)
b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)
Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)
(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)
a./
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)
(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Bạn chuyển về 1 vế sau đó trừ 1 vào mỗi phân thức ta được :
\(\left(x-2005\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005}\right)=0\)
Vì biểu thức bên phải khác 0 nên : \(x-2005=0\)=> \(x=2005\)
\(\frac{x-5}{2000}+\frac{x-4}{2001}+\frac{x-3}{2002}=\frac{x-2}{2003}+\frac{x-1}{2004}+\frac{x}{2005}\)
\(\Leftrightarrow\frac{x-2005}{2000}+\frac{x-2005}{2001}+\frac{x-2005}{2002}=\frac{x-2005}{2003}+\frac{x-2005}{2004}+\frac{x-2005}{2005}\)
\(\Leftrightarrow\left(x-2005\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005}\right)=0\)
<=> x - 2005 = 0
<=> x = 2005
Vậy ...............
a/ (X+1)/35+1+(x+3)/33+1 =(x+5)/31+(x+7)/29+1+1
=>(x+36)/35+(x+36)/33-(x+36)/31-(x+36)/27=0
=>(X+36)(1/35+1/33-1/31-1/29)=0
=> x+36=0(vì c=vế 2 luôn luôn khác 0)
=>x=-36
b/ CMTT câu a
trừ tung phân số cho 1 ta được x=2004
x= -2002 nhan. Dùng máy tính cầm tay sẽ ra
\(\frac{x+1}{2001}+\frac{x+2}{200}=\frac{x+3}{1999}+\frac{x+4}{1998}\)
\(\left(\frac{x+1}{2001}+1\right)+\left(\frac{x+2}{2000}+1\right)=\left(\frac{x+3}{1999}+1\right)+\left(\frac{x+4}{1998}+1\right)\)
\(\frac{x+2002}{2001}+\frac{x+2002}{2000}=\frac{x+2002}{1999}+\frac{x+2002}{1998}\)
\(\frac{x+2002}{2001}+\frac{x+2002}{2000}-\frac{x+2002}{1999}-\frac{x+2002}{1998}=0\)
\(\left(x+2002\right).\left(\frac{1}{2001}+\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
\(\Rightarrow x+2002=0\)
\(\Rightarrow x=0-2002\)
\(\Rightarrow x=-2002\)