`\text{~~Holi~~}`
$$\sqrt[3]{9-\sqrt{x+1}}+\sqrt[3]{7+\sqrt{x+1}}=4$$
Đặt `\sqrt{x+1}=t`. Khi đó phương trình trở thành:
$$\sqrt[3]{9-t}+\sqrt[3]{7+t}=4$$
`->` $$\sqrt[3]{9-t}=4-\sqrt[3]{7+t}$$
`->` $$9-t=64-48\sqrt[3]{7+t}+12\sqrt[3]{(7+t)^2}-(7+t)$$
`->` $$9-t=64-48\sqrt[3]{7+t}+12\sqrt[3]{(7+t)^2}-7-t$$
`->` $$9=64-48\sqrt[3]{7+t}+12\sqrt[3]{(7+t)^2}-7$$
`->` $$9=57-48\sqrt[3]{7+t}+12\sqrt[3]{(7+t)^2}$$
`->` $$9-57+48\sqrt[3]{7+t}-12\sqrt[3]{(7+t)^2}=0$$
`->` $$-48+48\sqrt[3]{7+t}-12\sqrt[3]{(7+t)^2}=0$$
`->` $$4-4\sqrt[3]{7+t}+\sqrt[3]{(7+t)^2}=0$$
`->` $$\sqrt[3]{(7+t)^2}-4\sqrt[3]{7+t}+4=0$$
Đặt $\sqrt[3]{7+t}=u$. Khi đó phương trình trở thành:
`u^2-4u+4=0`
`-> u=2`
`->` $\sqrt[3]{7+t}=2$
`-> t=1`
`->` $\sqrt{x+1}=1$
`-> x+1=1`
`-> x=0`
Vậy `S={0}`
