Question

Tính giá trị biểu thức \(\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\)với \(x=5+2\sqrt{7}\)

Answer 1

Đk: x = \(5+2\sqrt{7}\)> 5

Đặt A = \(\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\)

A² = \(\left(\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\right)^2\)

A² = \(3x+\sqrt{6x-1}+3x-\sqrt{6x-1}-2\sqrt{\left(3x+\sqrt{6x-1}\right)\left(3x-\sqrt{6x-1}\right)}\)

A² = \(6x-2\sqrt{9x^2-6x+1}\)

A² = \(6x-2\sqrt{\left(3x-1\right)^2}\) (vì x > \(\frac{1}{3}\))

A² = \(6x-2\left(3x-1\right)\)

A² = \(6x-6x+2\)

A² = 2

=> A = \(\sqrt{2}\)

Vậy ....

Answer 2

Đặt:    \(A=\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\)

=>    \(A^2=3x+\sqrt{6x-1}+3x-\sqrt{6x-1}-2\sqrt{\left(3x+\sqrt{6x-1}\right)\left(3x-\sqrt{6x-1}\right)}\)

=>    \(A^2=6x-2\sqrt{9x^2-6x+1}\)

=>    \(A^2=6x-2\sqrt{\left(3x-1\right)^2}\)

Mà:    \(x=5+2\sqrt{7}\Rightarrow x>\frac{1}{3}\Rightarrow3x>1\Rightarrow3x-1>0\)

=>   \(A^2=6x-2\left(3x-1\right)\)

=>    \(A^2=6x-6x+2=2\)

Mà:    \(\sqrt{3x+\sqrt{6x-1}}>\sqrt{3x-\sqrt{6x-1}}\Rightarrow A>0\)

=>    \(A=\sqrt{2}\)

VẬY    \(A=\sqrt{2}\)