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\(\frac{x+1}{2003}+\frac{x+3}{2001}+\frac{x+5}{1999}=\frac{x+7}{1997}+\frac{x+9}{1995}+\frac{x+11}{1993}\)
\(\Leftrightarrow\frac{x+1}{2003}+1+\frac{x+3}{2001}+1+\frac{x+5}{1999}+1=\frac{x+7}{1997}+1+\frac{x+9}{1995}+1+\frac{x+11}{1993}+1\)
\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}=\frac{x+2004}{1997}+\frac{x+2004}{1995}+\frac{x+2004}{1993}\)
\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}-\frac{x+2004}{1997}-\frac{x+2004}{1995}-\frac{x+2004}{1993}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2004=0\) ( do \(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\ne0\))
\(\Leftrightarrow x=-2004\)
\(\frac{x+1}{2003}\)\(+\)\(\frac{x+3}{2001}\)\(+\)\(\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+\frac{x+9}{1995}\)\(+\frac{x+11}{1993}\)
\(\Leftrightarrow\)\(\frac{x+1}{2003}\)\(+1+\)\(\frac{x+3}{2001}\)\(+1+\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+1+\frac{x+9}{1995}\)\(+1+\frac{x+11}{1993}\)
\(\Leftrightarrow\frac{x+2004}{2003}\)\(+\frac{x+2004}{2001}\)\(+\frac{x+2004}{1999}\)\(-\frac{x+2004}{1997}\)\(-\frac{x+2004}{1995}\)\(-\frac{x+2004}{1993}\)\(=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2004=0\)(vì tích kia có kết quả khác 0)
\(\Leftrightarrow x=-2004\)
Vậy PT có tập nghiệm S = {-2004}
a, \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
\(\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
\(\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
\(=>x+36=0\)
\(=>x=36\)
Giải phương trình sau:
\(\frac{x+2001}{5}+\frac{x+1999}{7}+\frac{x+1997}{9}+\frac{x+1995}{11}=-4.\)-4
\(\frac{x+2001}{5}+\frac{x+1999}{7}+\frac{x+1997}{9}+\frac{x+1995}{11}=-4\)
\(\Rightarrow\frac{x+2001}{5}+1+\frac{x+1999}{7}+1+\frac{x+1997}{9}+1+\frac{x+1995}{11}+1=0\)
\(\Rightarrow\frac{x+2006}{5}+\frac{x+2006}{7}+\frac{x+2006}{9}+\frac{x+2006}{11}=0\)
\(\Rightarrow\left(x+2006\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\right)=0\)
\(\Rightarrow x+2006=0\)vì \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}>0\)
\(\Rightarrow x=-2006\)
\(\frac{x+6}{1999}+\frac{x+8}{1997}=\frac{x+10}{1995}+\frac{x+12}{1993}\)
\(\Leftrightarrow\frac{x+6}{1999}+1+\frac{x+8}{1997}+1=\frac{x+10}{1995}+1+\frac{x+12}{1993}+1\)
\(\Leftrightarrow\frac{x+2005}{1999}+\frac{x+2005}{1997}=\frac{x+2005}{1995}+\frac{x+2005}{1993}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{1999}+\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2005=0\left(\frac{1}{1999}+\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\ne0\right)\)
<=> x=-2005
Vậy x=-2005
bạn chỉ cần cộng mỗi phân số với 1 là xong!
Vd: x+6/1999 +1 +x+8/1997 +1 = x+10/1995 +1 +x+12/1993 +1
(không quen sử dụng cái phần mềm này lắm nên mình không làm nốt được)
a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)
\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)
\(\Rightarrow x=-100\)
b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)
\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)
\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)
b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)
=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)
\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)
\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0
\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0
\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))
\(\Leftrightarrow\)x=-1999
Vậy x=-1999
a, \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1-\left(\frac{x+5}{61}+1\right)-\left(\frac{x+7}{59}+1\right)=0\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}\)=0
<=> \(\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
<=> x+66=0 \(\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)\)
<=> x=-66
Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)
\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)
\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)
\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)
\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)
Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)
=> x - 2000 = 0
=> x = 2000