Question

Answer 1

\(14x-42+13x-39=16x-48+15x-45\)

\(\Leftrightarrow14x+13x-16x-15x=42+39-48-45\)

\(\Leftrightarrow-4x=-12\)

\(\Rightarrow x=3\)

Answer 2

Ta có : \(\dfrac{x-3}{13}+\dfrac{x-3}{14}=\dfrac{x-3}{15}+\dfrac{x-3}{16}\)

\(\Leftrightarrow\dfrac{x-3}{13}+\dfrac{x-3}{14}-\dfrac{x-3}{15}-\dfrac{x-3}{16}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)

Vì \(\left(\dfrac{1}{13}+\dfrac{1}{14}-\dfrac{1}{15}-\dfrac{1}{16}\right)>0\)

⇒\(x-3=0\Leftrightarrow x=3\)