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

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

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

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

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

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

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

\(\Leftrightarrow x-1995=0\).Do \(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}+\frac{1}{15}+\frac{1}{25}\ne0\)

\(\Leftrightarrow x=1995\)

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

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

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

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

\(\Leftrightarrow x-1995=0\)

\(\Leftrightarrow x=1995\)

( 1308³.1990) . ( 1209⁵.56) . ( 1 + 2 + 3 + 4 + ... + 999 + 1000 ) .( 122.2 - 61.4 )

= ( 1308³.1990) . ( 1209⁵.56) . ( 1 + 2 + 3 + 4 + ... + 999 + 1000 ) . 0

= 0 

( 1308³.1990) . ( 1209⁵.56) . ( 1 + 2 + 3 + 4 + ... + 999 + 1000 ) .( 122.2 - 61.4 )

= ( 1308³.1990) . ( 1209⁵.56) . ( 1 + 2 + 3 + 4 + ... + 999 + 1000 ) . 0

= 0 

s: \(=125\cdot100-25\cdot192\)

\(=12500-4800\)

=7700

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

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

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

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

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

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

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

nên x-1995=0

hay x=1995

Vậy: S={1995}

a) Ta có:mx=2-x

⇔mx+x=2

⇔x(m+1)=2

Để phương trình mx=2-x vô nghiệm thì x(m+1)=0

⇔m+1=0

hay m=-1