cộng 1 vào mỗi tỉ số,ta đc:

(x+5)/1995+1+(x+4)/1996+1+(x+3)/1997+1=(x+1995)/5+1+(x+1996)/4+1+(x+1997|/3+1

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

\(\Rightarrow\frac{x+2000}{1995}+\frac{x+2000}{1996}+\frac{x+2000}{1997}-\frac{x+2000}{5}-\frac{x+2000}{4}-\frac{x-2000}{3}=0\)

\(\Rightarrow\left(x+2000\right)\left(\frac{1}{1995}+\frac{1}{1996}+\frac{1}{1997}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

mà bt trong ngoặc thứ 2 khác 0

=>x+2000=0

=>x=-2000

x=-2000           

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 )

                                                          \(\Rightarrow x=2000\)

(x-5)(1995.1996+1997.1996)=1234.5678.(630-315.2):1996

VP=1234.5678(630-315.2):196

=0

=> (x-5)(1995.1996+1997.1996)=0

=> x-5=0  vì (1995.1996+1997.1996 khác 0)

=> x=5

( x - 5 ) x ( 1995 x 1996 + 1997 x 1996 ) = 1234 x 5678 x ( 630 - 315 x 2 ) : 1996

=( x - 5 ) x ( 1995 x 1996 + 1997 x 1996 ) = 0

=x-5=0:( 1995 x 1996 + 1997 x 1996 )

=x-5=0

x=0+5

x=5

x = 5 vì vế kia bằng 0 thì nếu x = 5 mà 5 - 5 bằng 0 nếu 0 nhân với bất kỳ số nào cũng bằng 0

a) \(\dfrac{1997x1996-1}{1995x1997+1996}=\dfrac{1997x\left(1995+1\right)-1}{1995x1997+1996}\)

\(=\dfrac{1997x1995+1997-1}{1995x1997+1996}=\dfrac{1997x1995+1996}{1995x1997+1996}=1\)

b) \(\dfrac{1997x1996-995}{1995x1997+1002}=\dfrac{1997x\left(1995+1\right)-995}{1995x1997+1002}\)

\(=\dfrac{1997x1995+1997-995}{1995x1997+1002}=\dfrac{1997x1995+1002}{1995x1997+1002}=1\)

a) \(\frac{35}{14}=\frac{5\times7}{2\times7}=\frac{5}{2}\)

\(\frac{125}{50}=\frac{5\times5\times5}{2\times5\times5}=\frac{5}{2}\)

b)\(\frac{4\times5+4\times11}{8\times7+4\times3}=\frac{4\times\left(5+11\right)}{4\times\left(2\times7+3\right)}=\frac{16}{17}\)

c) \(\frac{3\times11+7\times11}{22\times2+11\times6}=\frac{11\times\left(3+7\right)}{22\times\left(2+3\right)}=\frac{11\times10}{22\times5}=\frac{11\times2\times5}{11\times2\times5}=1\)

(x-5) x ( 1995 x 196 + 1997 x 1996) = 0

x-5 = 0 : ( 1995 x 196 + 1997 x 1996)

x-5 = 0

x= 0+5

x= 5