1. \(\Leftrightarrow\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{51-x}{49}+1=-5+5\)

 \(\Leftrightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)

 \(\Leftrightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)

 \(\Leftrightarrow x-100=0\Leftrightarrow x=100\)

2. \(\Leftrightarrow\frac{x-5}{1990}+1+\frac{x-15}{1980}+1+\frac{x-25}{1970}=\frac{x-1990}{5}+1+\frac{x-1980}{15}+1+\frac{x-1970}{25}+1\)

   \(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)

   \(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)

  \(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)

  \(\Leftrightarrow x-1995=0\Leftrightarrow x=1995\)

câu 3 hình như sai đề

\(\Leftrightarrow\left(\dfrac{x-5}{1990}-1\right)+\left(\dfrac{x-15}{1980}-1\right)+\left(\dfrac{x-25}{1970}-1\right)\\ +\left(\dfrac{x-1990}{5}-1\right)+\left(\dfrac{x-1980}{15}-1\right)+\left(\dfrac{x-1970}{25}-1\right)=0\\ \Leftrightarrow\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}+\dfrac{x-1995}{1970}+\dfrac{x-1995}{5}\\ +\dfrac{n-1995}{15}+\dfrac{n-1995}{25}=0\\ \Rightarrow\left(x-1995\right)\left(\dfrac{1}{1990}+\dfrac{1}{1980}+\dfrac{1}{1970}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}\right)=0\) 

\(\Rightarrow x-1995=0\\ \Rightarrow x=1995\)

xong r nhé. thanks m.n

Ta có: \(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}\)

\(\Leftrightarrow\)\(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}-3=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}-3\)

\(\Leftrightarrow\)\(\frac{x-5}{1990}-1+\frac{x-15}{1980}-1+\frac{x-25}{1970}-1=\frac{x-1990}{5}-1+\frac{x-1980}{15}-1+\frac{x-1970}{25}-1\)\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)

\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)

\(\Leftrightarrow\)\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)

\(\Leftrightarrow\)\(x-1995=0\)

\(\Leftrightarrow\)\(x=1995\)

\(\frac{1+0,6-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{\frac{3}{3}+\frac{3}{5}-\frac{3}{7}}{\frac{8}{3}+\frac{8}{5}-\frac{8}{7}}=\frac{3.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}{8.\left(\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\right)}=\frac{3.1}{8.1}=\frac{3}{8}\)

\(\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}=\frac{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}}{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}+\frac{7}{16}}=\frac{1.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}{7.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{8}\right)}=\frac{1.1}{7.1}=\frac{1}{7}\)

=>\(\frac{3}{8}-\frac{1}{7}=\frac{13}{56}\)

Mk đg cần gấp các bn giúp mk nha

Cảm ơn bạn nha,  Nguyễn Thành An

E = ( x - 29 ) / 1970 + ( x - 27 ) / 1972 + ( x - 25 ) / 1974 + ( x - 23 ) / 1976 + ( x - 21 ) / 1978 + ( x - 19 ) / 1980 = ( x - 1970 ) / 29 + ( x - 1972 ) / 27 + ( x - 1974 ) / 25 + ( x - 1976 ) / 23 + ( x - 1978 ) / 21 + ( x - 1980 ) / 19

( Trừ từng số hạng cho 1 ra như sau )

E = (x - 1999)/ 1970 + ( x - 1999 ) / 1972 + ( x - 1999) / 1974 + ( x - 1999)/ 1976 + ( x -1999) / 1978 + ( x - 1999)/ 1980 = ( x - 1999)/29 + ( x - 1999) / 27 + ( x - 1999 ) / 25 + ( x - 1999) / 23 + ( x - 1999)/21 + ( x - 1999) / 19

< = > ( x - 1999 ) / 1970 + ( x - 1999 ) / 1972 + ( x - 1999 ) / 1974 + ( x - 1999) / 1976 + ( x - 1999) / 1978 + ( x - 1999) / 1980 - ( x - 1999) / 29 - ( x - 1999)/ 27 - ( 1 - 1999) / 25 - ( x-1999) / 23 - ( x - 1999) / 21 - ( x - 1999) / 19 = 0 ( chuyển vế )

< = > ( x - 1999 ) ( 1/1970 + 1/ 1972 + 1/1974 + 1/1976 + 1/1978 + 1/1980 - 1/29 - 1/27 - 1/25 - 1/23 - 1/21 - 1/19) = 0

Vì ( 1/1970 + 1/1972 + 1/1974 + 1/1976 + 1/1978 + 1/1980 - 1/29 -1/27 - 1/25 - 123 - 1/21 - 1/19 ) khác 0 nên để đẳng thức bằng 0 thì bắt buộc x - 1999 = 0

< = > x = 0 + 1999 = 1999

Vậy tập nghiệm của phương trình là S = { 1999 }

pt <=> (x-5/1990 -  1) + (x-15/1980 - 1) = (x-1980/15 - 1) + (x-1990/5 - 1)

<=> x-1995/1990 + x-1995/1980 = x-1995/15 + x-1995/5

<=> x-1995/15 + x-1995/5 - x-1995/1990 - x-1995/1980 = 0

<=> (x-1995).(1/5+1/15-1/1990-1/1980) = 0

<=> x-1995 = 0 ( vì 1/5 + 1/15 - 1/1990 - 1/1980 > 0 )

<=> x = 1995

Vậy S={1995}

Tk mk nha

Ta có : 

\(\frac{x-5}{1990}+\frac{x-15}{1980}=\frac{x-1980}{15}+\frac{x-1990}{5}\)

\(\Leftrightarrow\)\(\left(\frac{x-5}{1990}-1\right)+\left(\frac{x-15}{1980}-1\right)=\left(\frac{x-1980}{15}-1\right)+\left(\frac{x-1990}{5}-1\right)\)

\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}=\frac{x-1995}{15}+\frac{x-1995}{5}\)

\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)

\(\Leftrightarrow\)\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{15}+\frac{1}{5}\right)=0\)

Vì \(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{15}+\frac{1}{5}\ne0\)

Nên \(x-1995=0\)

\(\Rightarrow\)\(x=1995\)

Vậy \(x=1995\)

Chúc bạn học tốt ~

\(\Leftrightarrow\left(\frac{x-5}{1990}-1\right)+\left(\frac{x-15}{1980}-1\right)=\left(\frac{x-1980}{15}-1\right)+\left(\frac{x-1990}{5}-1\right)\)

\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}=\frac{x-1995}{15}+\frac{x-1995}{5}\)

\(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\right)=0\)

\(\Leftrightarrow x-1995=0\)

\(\Leftrightarrow x=1995\)

Vậy pt có No Ià x=1995