Question

rút gọn \(B=\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)

Answer 1

Áp dụng hằng đẳng thức (a+b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab.(a +b) ta có: 

\(B^3=20+14\sqrt{2}+20-14\sqrt{2}+3\sqrt[3]{20+14\sqrt{2}}.\sqrt[3]{20-14\sqrt{2}}\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)\)

\(B^3=40+3\sqrt[3]{\left(20+14\sqrt{2}\right)\left(20-14\sqrt{2}\right)}.B\)

\(B^3=40+3.\sqrt[3]{400-392}.B=40+3.\sqrt[3]{8}.B=40+6B\)

=> B³ - 6B - 40 = 0

<=> B³ - 64 - 6B + 24  = 0

<=> (B - 4 ).(B² + 4B + 16) - 6.(B - 4) = 0

<=> (B - 4).(B² + 4B + 16 - 6) = 0 <=> B = 4 hoặc B² + 4B + 10 = 0

B² + 4B + 10 = 0 Vô nghiêm vì \(\Delta\) = 16 - 40 = -24 < 0

Vậy B = 4