\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)
\(\Leftrightarrow12x^2-24=2x\)
\(\Leftrightarrow12x^2-2x-24=0\)
\(\text{Δ}=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156>0\)
DO đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{2-34}{24}=\dfrac{-32}{24}=\dfrac{-4}{3}\left(nhận\right)\\x_2=\dfrac{2+34}{24}=\dfrac{36}{24}=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
`x/2-1/x=1/12`
`=>[x.6x]/[12x]-12/[12x]=x/[12x]`
`=>6x^2-12=x`
`=>6x^2-x-12=0`
`=>6x^2-9x+8x-12=0`
`=>3x(2x-3)+4(2x-3)=0`
`=>(2x-3)(3x+4)=0`
`@TH1:2x-3=0=>2x=3=>x=3/2`
`@TH2:3x+4=0=>3x=-4=>x=-4/3`
