\(\dfrac{4}{x}+\dfrac{4}{x+3}=\dfrac{3}{5}\)
`<=>`\(\dfrac{4\cdot\left(x+3\right)+4\cdot x}{x\cdot\left(x+3\right)}=\dfrac{3}{5}\)
`<=>`\(\dfrac{4x+12+4x}{x\cdot\left(x+3\right)}=\dfrac{3}{5}\)
`<=>`\(\dfrac{\left(4x+4x\right)+12}{x\cdot\left(x+3\right)}=\dfrac{3}{5}\)
`<=>`\(\dfrac{8x+12}{x\cdot\left(x+3\right)}=\dfrac{3}{5}\)
`<=>`\(\dfrac{8x+12}{x\cdot x+x\cdot3}=\dfrac{3}{5}\)
`<=>`\(\dfrac{8x+12}{x^2+3x}=\dfrac{3}{5}\)
`<=>`\(\left(8x+12\right)\cdot5=3\cdot\left(x^2+3x\right)\)
`<=>`\(40x+60=3x^2+9x\)
`<=>`\(40x+60-3x^2-9x=0\)
`<=>`\(\left(40x-3x^2-9x\right)+60=0\)
`<=>`\(-3x^2+31x+60=0\)
`<=>`\(-\left(3x^2+5x-31x-60\right)=0\)
`<=>`\(-\left(x\cdot3x+5\right)-12\left(3x+5\right)=0\)
`<=>`\(-\left(3x+5\right)\cdot\left(x-12\right)=0\)
`<=>`\(\left[{}\begin{matrix}3x+5=0\\x-12=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=12\end{matrix}\right.\)
Mất mấy chục phút để làm đó :v
Đk: `x ne 0, -3`.
`-> (4x + 12 + 4x)/(x(x+3)) = 3/5`
`-> (8x + 12)/(x^2+3x) = 3/5`
`-> 40x + 60 = 3x^2 + 9x`
`-> 3x^2 - 31x - 60 = 0`
`-> x = -5/3` hoặc `x = 12`
