Question

\(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\) a)
\(\dfrac{x-3}{x-2}+\dfrac{3-2}{x-4}=3\dfrac{1}{5}\) b)
\(3x^2+2x-1=0\) c)

Answer 1

a. \(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}=\dfrac{x+100}{94}+\dfrac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

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

c. \(\Leftrightarrow3x^2+3x-x-1=0\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow\left[\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)