Question

Cho biểu thức A= \(\frac{\sqrt{x}+4}{\sqrt{x}+2}\) tính giá trị của A khi x=3

Answer 1

\(A=\frac{\sqrt{3}+4}{\sqrt{3}+2}\)

\(A=\frac{\sqrt{3}+2}{\sqrt{3}+2}+\frac{2}{\sqrt{3}+2}\)

\(A=1+\frac{2}{\sqrt{3}+2}\)

\(A=1+\frac{2\left(2-\sqrt{3}\right)}{\left(\sqrt{3}+2\right)\left(2-\sqrt{3}\right)}\)

\(A=1+\frac{4-2\sqrt{3}}{4-3}\)

\(A=1+4-2\sqrt{3}\)

\(A=5-2\sqrt{3}\)

Answer 2

+ với x = 3 ta có :

\(A=\frac{\sqrt{3}+4}{\sqrt{3}+2}=\frac{5\sqrt{3}+10-4\sqrt{3}-6}{\sqrt{3}+2}\) \(=\frac{5\left(\sqrt{3}+2\right)-2\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}+2}\)

\(=\frac{\left(5-2\sqrt{3}\right)\left(\sqrt{3}+2\right)}{\sqrt{3}+2}=5-2\sqrt{3}\)