`B=({x-\sqrt{x}}/{\sqrt{x}-1}-{\sqrt{x}+1}/{x+\sqrt{x}}):{\sqrt{x}+1}/{\sqrt{x}}` $(x>0;x\ne 1)$
`B=[{\sqrt{x}(\sqrt{x}-1)}/{\sqrt{x}-1}-{\sqrt{x}+1}/{\sqrt{x}(\sqrt{x}+1)}].{\sqrt{x}}/{\sqrt{x}+1}`
`B=(\sqrt{x}-1/{\sqrt{x}}).{\sqrt{x}}/{\sqrt{x}+1}`
`B={x-1}/{\sqrt{x}}. {\sqrt{x}}/{\sqrt{x}+1}`
`B={(\sqrt{x}+1)(\sqrt{x}-1)}/{\sqrt{x}+1}`
`B=\sqrt{x}-1`
$\\$
`x=12+8\sqrt{2}=(2\sqrt{2})^2+2.2\sqrt{2}.2+2^2`
`x=(2\sqrt{2}+2)^2`
Thay `x=(2\sqrt{2}+2)^2` vào `B` ta có:
`B=\sqrt{x}-1=\sqrt{(2\sqrt{2}+2)^2}-1`
`B=2\sqrt{2}+2-1=2\sqrt{2}+1`
Vậy `B=2\sqrt{2}+1`