TA CÓ X = √3 20+14√2+√320−14√2⇒X3=40+3√3(20+14√2)(20−14√2)(√320+14√2...

Câu 3: Ta có x = √

3

20+14

2+

3

2014

2⇒x

3

=40+3

3

(20+14

2)(20−14

2)(

3

20+14

2+

3

20−14

2)=40+6x

 x

3

– 6x = 40 => f(x) = 40.