Chứng minh nghiệm của phương trình sau là một số nguyên:
\(\frac{x-2}{2005}\) + \(\frac{x-3}{2004}\) + \(\frac{x-4}{2003}\) = \(\frac{x-2005}{2}\) + \(\frac{x-2004}{3}\) + \(\frac{x-2003}{4}\)
Giúp mình nha mình đang vội lắm!!!
Giải phương trình sau:
\(\frac{x+2}{2005}+\frac{x+3}{2004}+\frac{x+4}{2003}+3=0\)
Bài làm
\(\frac{x+2}{2005}+\frac{x+3}{2004}+\frac{x+4}{2003}+3=0\)
\(\Leftrightarrow\left(\frac{x+2}{2005}+1\right)+\left(\frac{x+3}{2004}+1\right)+\left(\frac{x+4}{2003}+1\right)=0\)
\(\Leftrightarrow\left(\frac{x+2+2005}{2005}\right)+\left(\frac{x+3+2004}{2004}\right)+\left(\frac{x+4+2003}{2003}\right)=0\)
\(\Leftrightarrow\frac{x+2007}{2005}+\frac{x+2007}{2004}+\frac{x+2007}{2003}=0\)
\(\Leftrightarrow\left(x+2007\right).\frac{1}{2005}+\left(x+2007\right).\frac{1}{2004}+\left(x+2007\right).\frac{1}{2003}=0\)
\(\Leftrightarrow\left(x+2007\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2007=\frac{0}{\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}}\)
\(\Leftrightarrow x+2007=0\)
\(\Leftrightarrow x=-2007\)
Vậy phương trình trên có tập nghiệm S = { -2007 }
# Học tốt #
\(\frac{x+2}{2005}+\frac{x+3}{2004}+\frac{x+4}{2003}+3=0\)
\(\Leftrightarrow\left(\frac{x+2}{2005}+1\right)+\left(\frac{x+3}{2004}+1\right)+\left(\frac{x+4}{2003}+1\right)=0\)
\(\Leftrightarrow\frac{x+2007}{2005}+\frac{x+2007}{2004}+\frac{x+2007}{2003}=0\)
\(\Leftrightarrow\left(x+2007\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)(1)
Vì \(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}>0\)(2)
Từ (1), (2) \(\Rightarrow x+2017=0\)\(\Leftrightarrow x=-2017\)
Vậy \(x=-2017\)
Tìm x: a, \(\frac{x-2004}{2003}+\frac{x-2003}{2004}+\frac{x-2005}{2004}=3+\frac{2005}{2003}\)\(+\frac{2004}{2005}\)
c) 22/5 + 51/9 + 11/4 + 3/5 + 1/3 + 1/4
= 22/5 +3/5 +51/9 + 1/3 +11/4+1/4
= (22/5 +3/5) +(51/9 + 3/9) +(11/4+1/4)
= 25/5 +54/9 +12/4
= 5 +6 +3
= 14
d) (1/6 + 1/10 + 1/15) : (1/6 + 1/10 - 1/15)
= (5/30 + 3/30 +2/30 ) :(5/30 +3/30 -2/30)
= 10/30 : 6/30
= 1/3 : 1/5
= 5/3
Giải phương trình sau :
\(\frac{x^2-2008}{2007}+\:\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\:\frac{x^2-\:2005}{2004}+\:\frac{x^2-2004}{2003}+\:\frac{x^2-2003}{2002}\)
Ta có : \(\frac{x^2-2008}{2007}+\frac{x^2-2007}{2006}+\frac{x^2-2006}{2005}=\frac{x^2-2005}{2004}+\frac{x^2-2004}{2003}+\frac{x^2-2003}{2002}\)
=> \(\frac{x^2-2008}{2007}+1+\frac{x^2-2007}{2006}+1+\frac{x^2-2006}{2005}+1=\frac{x^2-2005}{2004}+1+\frac{x^2-2004}{2003}+1+\frac{x^2-2003}{2002}+1\)
=> \(\frac{x^2-2008}{2007}+\frac{2007}{2007}+\frac{x^2-2007}{2006}+\frac{2006}{2006}+\frac{x^2-2006}{2005}+\frac{2005}{2005}=\frac{x^2-2005}{2004}+\frac{2004}{2004}+\frac{x^2-2004}{2003}+\frac{2003}{2003}+\frac{x^2-2003}{2002}+\frac{2002}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}=\frac{x^2-1}{2004}+\frac{x^2-1}{2003}+\frac{x^2-1}{2002}\)
=> \(\frac{x^2-1}{2007}+\frac{x^2-1}{2006}+\frac{x^2-1}{2005}-\frac{x^2-1}{2004}-\frac{x^2-1}{2003}-\frac{x^2-1}{2002}=0\)
=> \(\left(x^2-1\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
=> \(x^2-1=0\)
=> \(x^2=1\)
=> \(x=\pm1\)
Vậy phương trình có 2 nghiệm là x = 1, x = -1 .
giải phương trình sau :
\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
dễ mà bn,cộng 1 vào mỗi biểu thức và trừ vế 2 là xong
\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+3=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}+3\)
\(\Leftrightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)\)
\(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)(1)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\)(2)
Từ (1) và (2) \(\Rightarrow x+2009=0\)\(\Rightarrow x=-2009\)
Vậy \(x=-2009\)
\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+3=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}+3\)
\(\Leftrightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+2009\right)\left(\frac{1}{2005}+\frac{2}{2004}+\frac{1}{2003}\right)\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Rightarrow x+2009=0\)( Vì \(\frac{1}{2008}+\frac{1}{207}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\))
=> x = -2009
\(\frac{x-5}{2000}+\frac{x-4}{2001}+\frac{x-3}{2002}=\frac{x-2}{2003}+\frac{x-1}{2004}+\frac{x}{2005}\)
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 ...............
Giải phương trình:
1. \(\left(x^2+x\right)^2+4\left(x^2+x\right)^2=12\)
2. \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
câu 2 :
\(\Leftrightarrow\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}-\frac{x+4}{2005}-\frac{x+5}{2004}-\frac{x+6}{2003}\)=0
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x-2009}{2003}\)=0
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)
\(\Rightarrow x+2009=0\)
\(\Rightarrow x=-2009\)
1.
a.Tính : P = \(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
b.Biết : 13 + 23 + ... + 103 = 3025. Tính S = 23 + 43 + 63 + ... + 203
c.Cho A=\(\frac{x^3-3x^2+0,25xy^2-4}{x^2+y}\)Tính giá trị của A biết x=\(\frac{1}{2}\), y là số nguyên âm lớn nhất
2.Một con thỏ chạy trên một con đường mà hai phần ba con đường băng qua đồng cỏ và đoạn đường còn lại đi qua đầm lầy. Thời gian con thỏ chạy trên đồng cỏ bằng nửa thời gian chạy qua đầm lầy. Hỏi vận tốc của con thỏ trên đoạn đường nào lớn hơn ? Tính tỉ số vận tốc của con thỏ trên hai đoạn đường ?
tìm x biết :\(\frac{x-1}{2005}+\frac{x-2}{2004}-\frac{x-3}{2003}=\frac{x-4}{2002}\) Giúp mình với
\(\frac{x-1}{2005}+\frac{x-2}{2004}-\frac{x-3}{2003}=\frac{x-4}{2002}\)
=>\(\frac{x-1}{2005}+\frac{x-2}{2004}-\frac{x-3}{2003}-\frac{x-4}{2004}=0\)
=>\(\left(\frac{x-1}{2005}-1\right)+\left(\frac{x-2}{2004}-1\right)-\left(\frac{x-3}{2003}-1\right)-\left(\frac{x-4}{2002}-1\right)=0\)
=>\(\frac{x-1-2005}{2005}+\frac{x-2-2004}{2004}-\frac{x-3-2003}{2003}-\frac{x-4-2002}{2002}=0\)
=>\(\frac{x-2006}{2005}+\frac{x-2006}{2004}-\frac{x-2006}{2003}-\frac{x-2006}{2002}=0\)
=>\(\left(x-2006\right)\left(\frac{1}{2005}+\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
Mà \(\frac{1}{2005}+\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\ne0\)
=> x - 2006 = 0 => x = 2006
\(\frac{x+6}{2001}+\frac{x+5}{2002}+\frac{x+4}{2003}=\frac{x+3}{2004}+\frac{x+2}{2005}\)+\(\frac{x+1}{2006}\)
Tìm x