Đặt $AB^2=x$ $(x>0)$
Phương trình trở thành:
`\qquad 12={(9x-12)(12-x)}/{16}`
`<=>(9x-12)(12-x)=12.16=192`
`<=>108x-9x^2+12x-144-192=0`
`<=>-9x^2+120x-336=0`
$⇔\left[\begin{array}{l}x=4\\x=\dfrac{28}{3}\end{array}\right.$
+) `x=4<=>AB^2=4=>AB=2` (vì $AB>0$)
+) `x={28}/3<=>AB^2={28}/3`
`=>AB=\sqrt{{28}/3}=2\sqrt{7/3}={2\sqrt{21}}/3`
Vậy $AB=2$ hoặc `AB={2\sqrt{21}}/3`