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

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

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

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

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

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

Vì \(\frac{1}{1990}+\frac{1}{1980}\ne\frac{1}{15}+\frac{1}{5}\)           =>   \(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\ne0\)

=> x - 1995 = 0

=> x = 1995

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

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

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

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

<=>x=1995 

em mới học lớp 6 nha anh !!! ^^

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

x-5/1990+x-15/1980+x-25/1970=x-1990/5+x-1980/15+x-1970/25

<=> (x-5/1990-1)+(x-15/1980-1)+(x-25/1970-1)=(x-1990/5-1)+(x-1980/15-1)+(x-1970/25-1)

<=> x-1995/1990+x-1995/1980+x-1995/1970=x-1995/5+x-1995/15+x-1995/25

<=> (x-1995)(1/1990+1/1980+1/1970-1/5-1/15-1/25)=0

<=> x-1995=0 

<=> x=1995

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

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

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

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

Mk đg cần gấp các bn giúp mk nha

Cảm ơn bạn nha,  Nguyễn Thành An

\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}=\dfrac{x-1990}{5}+\dfrac{x-1980}{15}\\ =>\dfrac{x-5}{1990}-1+\dfrac{x-15}{1980}-1=\dfrac{x-1990}{5}-1+\dfrac{x-1980}{15}-1\\ =>\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}-\dfrac{x-1995}{5}-\dfrac{x-1995}{15}=0\\ =>\left(x-1995\right).\left(\dfrac{1}{1990}+\dfrac{1}{1980}-\dfrac{1}{5}-\dfrac{1}{15}\right)=0\\ =>x-1995=0\\ =>x=1995\)

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

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