Question

chứng minh đẳng thức

\(\sqrt[3]{6+\sqrt{\frac{847}{27}}}+\sqrt[3]{6-\sqrt{\frac{847}{27}}}=3\)

Answer 1

đặt \(a=\sqrt[3]{6+\sqrt{\frac{847}{27}}};b=\sqrt[3]{6-\sqrt{\frac{847}{27}}}\). dễ thấy a> 0; b > 0

=> \(a^3+b^3=6+\sqrt{\frac{847}{27}}+6-\sqrt{\frac{847}{27}}=12\); \(a.b=\sqrt[3]{6+\sqrt{\frac{847}{27}}}.\sqrt[3]{6-\sqrt{\frac{847}{27}}}=\sqrt[3]{36-\frac{847}{27}}=\frac{5}{3}\)

Có: (a+ b)³ = a³ + b³ + 3ab (a+ b)

=> (a + b)³ = 12 + 3. \(\frac{5}{3}\).(a + b) = 12+ 5.(a + b)

=> (a + b)³ - 5.(a +b)  - 12 = 0 

<=> (a + b)³ - 9.(a + b)  + 4.(a + b) - 12 = 0

<=> (a + b). [(a + b)² - 9] + 4.(a + b - 3) = 0 <=> (a + b).(a + b + 3).(a + b- 3) + 4.(a + b - 3) = 0 

<=> (a+ b - 3).[(a + b)(a+ b+ 3) + 4] = 0

<=> a+ b = 3 hoặc (a + b)(a+ b+ 3) + 4 = 0 

tuy nhiên : Vì a > 0; b > 0 nên (a + b)(a+ b+ 3) + 4 > 0 

vậy a + b = 3 => điều phải chứng minh