`a,` `B = ({2\sqrt[x]}/{\sqrt[x]+3}+{\sqrt[x]}/{\sqrt[x]-3}-{3x+3}/{x-9})÷({2\sqrt[x]-2}/{\sqrt[x]-3}-1)`
`B = {2x-6\sqrt[x]+x+3\sqrt[x]-3x-3}/{(\sqrt[x]+3)÷{2\sqrt[x]-2-\sqrt[x]+2}/{\sqrt[x]-2}`
`B = -3(\sqrt[x] + 1)/{(\sqrt[x]-3)(\sqrt[x]+3)} × {\sqrt[x] - 3}/{\sqrt[x] + 1}`
`B = {-3}/{\sqrt[x] + 3}`
`b` `(x ≥ 0;x \ne 9)`
Để `B ≤ -1/3`
`⇔ {-3}/{\sqrt[x] + 3} ≤ -1/3`
`⇔ {-3}/{\sqrt[x] + 3} - 1/3 ≥ 0`
`⇔ {-9\sqrt[x]-3}/{\sqrt[x] + 3} ≥ 0`
`⇔ 6 - \sqrt[x] ≥ 0`
`⇔ x ≤ 36`
Kết luận ta đc : `0 ≤ x ≤ 36, x \ne 9`
Xin hay nhất !
