Question

Cho P=\(\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\) .Tính giá trị của P biết \(x=7-4\sqrt{3}\) 

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Answer 1

\(P=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}=\dfrac{7-4\sqrt{3}+\sqrt{7-4\sqrt{3}}}{3\sqrt{7-4\sqrt{3}}-1}=\dfrac{7-4\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}}{3\sqrt{\left(2-\sqrt{3}\right)^2}-1}=\dfrac{7-4\sqrt{3}+\left|2-\sqrt{3}\right|}{3\left|2-\sqrt{3}\right|-1}=\dfrac{7-4\sqrt{3}+2-\sqrt{3}}{3\left(2-\sqrt{3}\right)-1}=\dfrac{9-5\sqrt{3}}{5-3\sqrt{3}}=\dfrac{\left(9-5\sqrt{3}\right)\left(5+3\sqrt{3}\right)}{\left(5-3\sqrt{3}\right)\left(5+3\sqrt{3}\right)}=\dfrac{45+2\sqrt{3}-45}{-2}=-\sqrt{3}\)

Answer 2

Thay \(x=7-4\sqrt{3}\) vào P, ta được:

\(P=\dfrac{7-4\sqrt{3}+2-\sqrt{3}}{6-3\sqrt{3}-1}\)

\(=\dfrac{9-5\sqrt{3}}{5-3\sqrt{3}}=-\sqrt{3}\)