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
