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17 tháng 8 2015

=> \(\frac{x-1}{1995}+1-1-\frac{x+3}{1991}=\frac{x+7}{1987}+1-1-\frac{x+11}{1983}\)

=> \(\left(\frac{x-1}{1995}+1\right)-\left(1+\frac{x+3}{1991}\right)=\left(\frac{x+7}{1987}+1\right)-\left(1+\frac{x+11}{1983}\right)\)

=> \(\frac{x+1994}{1995}-\frac{x+1994}{1991}=\frac{x+1994}{1987}-\frac{x+1994}{1983}\)

=> \(\left(x+1994\right)\left(\frac{1}{1995}-\frac{1}{1991}-\frac{1}{1987}+\frac{1}{1983}\right)=0\)

=>x + 1994 = 0 Vì \(\left(\frac{1}{1995}-\frac{1}{1991}-\frac{1}{1987}+\frac{1}{1983}\right)\ne0\)

=> x  = -1994

27 tháng 2 2020

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

2 tháng 4 2020

\(\frac{x+1}{2003}+\frac{x+3}{2001}+\frac{x+5}{1999}=\frac{x+7}{1997}+\frac{x+9}{1995}+\frac{x+11}{1993}\)

\(\Leftrightarrow\frac{x+1}{2003}+1+\frac{x+3}{2001}+1+\frac{x+5}{1999}+1=\frac{x+7}{1997}+1+\frac{x+9}{1995}+1+\frac{x+11}{1993}+1\)

\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}=\frac{x+2004}{1997}+\frac{x+2004}{1995}+\frac{x+2004}{1993}\)

\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}-\frac{x+2004}{1997}-\frac{x+2004}{1995}-\frac{x+2004}{1993}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\right)=0\)

\(\Leftrightarrow x+2004=0\) ( do \(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\ne0\))

\(\Leftrightarrow x=-2004\)

2 tháng 4 2020

\(\frac{x+1}{2003}\)\(+\)\(\frac{x+3}{2001}\)\(+\)\(\frac{x+5}{1999}\)\(\frac{x+7}{1997}\)\(+\frac{x+9}{1995}\)\(+\frac{x+11}{1993}\)

\(\Leftrightarrow\)\(\frac{x+1}{2003}\)\(+1+\)\(\frac{x+3}{2001}\)\(+1+\frac{x+5}{1999}\)\(\frac{x+7}{1997}\)\(+1+\frac{x+9}{1995}\)\(+1+\frac{x+11}{1993}\)

\(\Leftrightarrow\frac{x+2004}{2003}\)\(+\frac{x+2004}{2001}\)\(+\frac{x+2004}{1999}\)\(-\frac{x+2004}{1997}\)\(-\frac{x+2004}{1995}\)\(-\frac{x+2004}{1993}\)\(=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)

\(\Leftrightarrow x+2004=0\)(vì tích kia có kết quả khác 0)

\(\Leftrightarrow x=-2004\)

Vậy PT có tập nghiệm S = {-2004}

20 tháng 5 2019

Giải phương trình sau:
\(\frac{x-2004}{15}\)+\(\frac{x-1995}{12}\)+\(\frac{x-1989}{10}\)+\(\frac{x-1987}{8}\)=\(10\)

\(\frac{\left(x-2004\right).40}{600}\) +\(\frac{\left(x-1995\right).50}{600}\)+\(\frac{\left(x-1989\right).60}{600}\)+\(\frac{\left(x-1987\right).75}{600}\)=\(\frac{10.600}{600}\)

\(\frac{40x-80160}{600}\) + \(\frac{50x-99750}{600}\) +\(\frac{60x-119340}{600}\) +\(\frac{75x-149025}{600}\)=\(\frac{6000}{600}\)

\(40x-80160+50x-99750+60x-119340+75x-149025=6000\)\(225x=\)\(6000+80160+99750+119340+149025\)

\(225x=454275\)

\(x=2019\)

20 tháng 5 2019

Hỏi đáp Toán

.....

8 tháng 9 2016

\(\frac{x-1986-1987}{1985}+\frac{x-1985-1987}{1986}+\frac{x-1985-1986}{1987}=3\)

=> \(\left(\frac{x-1986-1987}{1985}-1\right)+\left(\frac{x-1985-1987}{1986}-1\right)+\left(\frac{x-1985-1986}{1987}-1\right)=3-3\)

=> \(\frac{x-1985-1986-1987}{1985}+\frac{x-1985-1986-1987}{1986}+\frac{x-1985-1986-1987}{1987}=0\)

=> \(\left(x-1985-1986-1987\right).\left(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}\right)=0\)

=> \(\left(x-5958\right).\left(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}\right)=0\)

Mà \(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}\ne0\)

=> x - 5958 = 0

=> x = 5958

x=-2000           

3 tháng 8 2015

ta có \(1+\frac{x+5}{1995}+1+\frac{x+4}{1996}+1+\frac{x+3}{1997}=1+\frac{x+1995}{5}+1+\frac{x+1996}{4}+1+\frac{x+1997}{3}\)

        \(=\frac{x+2000}{1995}+\frac{x+2000}{1996}+\frac{x+2000}{1997}=\frac{x+2000}{5}+\frac{x+2000}{4}+\frac{x+2000}{3}\)

     \(=\left(x+2000\right)\left(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\right)=\left(x+2000\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\)  (1)

                     Xét     \(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}\ne\frac{1}{5}+\frac{1}{4}+\frac{1}{3}vàx+2000=x+2000\) (2)

                                        từ \(\left(1\right)\Leftrightarrow x+2000=0\) ( để (1) là đúng )

                                                          \(\Rightarrow x=2000\)

Giải các phương trình sau : ( biến đổi đặc biệt )a) \(\frac{x+1}{35}\)+ \(\frac{x+3}{33}\)= \(\frac{x+5}{31}\)+ \(\frac{x+7}{29}\)( HD : cộng thêm 1 vào các hạng tử )b) \(\frac{x-10}{1994}\)+ \(\frac{x-8}{1996}\)+\(\frac{x-6}{1998}\)+ \(\frac{x-4}{2000}\)+ \(\frac{x-2}{2002}\)= \(\frac{x-2002}{2}\)+ \(\frac{x-2000}{4}\)+ \(\frac{x-1988}{6}\)+ \(\frac{x-1996}{8}\)+ \(\frac{x-1994}{10}\)( HD : trừ đi 1 vào các hạng tử...
Đọc tiếp

Giải các phương trình sau : ( biến đổi đặc biệt )

a) \(\frac{x+1}{35}\)\(\frac{x+3}{33}\)\(\frac{x+5}{31}\)\(\frac{x+7}{29}\)( HD : cộng thêm 1 vào các hạng tử )

b) \(\frac{x-10}{1994}\)\(\frac{x-8}{1996}\)+\(\frac{x-6}{1998}\)\(\frac{x-4}{2000}\)\(\frac{x-2}{2002}\)\(\frac{x-2002}{2}\)\(\frac{x-2000}{4}\)\(\frac{x-1988}{6}\)\(\frac{x-1996}{8}\)\(\frac{x-1994}{10}\)( HD : trừ đi 1 vào các hạng tử ) 

c) \(\frac{x-1991}{9}\)\(\frac{x-1993}{7}\)\(\frac{x-1995}{5}\)\(\frac{x-1997}{3}\)\(\frac{x-1991}{1}\)\(\frac{x-9}{1991}\)\(\frac{x-7}{1993}\)\(\frac{x-5}{1995}\)\(\frac{x-3}{1997}\)\(\frac{x-1}{1999}\)( HD : trừ đi 1 vào các hạng tử )

d) \(\frac{x-85}{15}\)\(\frac{x-74}{13}\)\(\frac{x-67}{11}\)\(\frac{x-64}{9}\)= 10  ( Chú ý : 10 = 1 + 2 + 3 + 4 )

e) \(\frac{x-1}{13}\)\(\frac{2x-13}{15}\)\(\frac{3x-15}{27}\)\(\frac{4x-27}{29}\)( HD : Thêm hoặc bớt 1 vào các hạng tử )

 

1
16 tháng 4 2020

a, \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)

\(\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)

\(\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)

\(=>x+36=0\)

\(=>x=36\)