Question

VD3 : Tính giá trị của các biểu thức sau

\(\text{e) }\frac{\left(2+\sqrt{3}\right)\left(\sqrt{2-\sqrt{3}}\right)}{\sqrt{2+\sqrt{3}}}\)

Answer 1

kết bạn nha

Answer 2

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

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

\(=\sqrt{4-3}=1\)

Answer 3

\(e,\frac{\left(2+\sqrt{3}\right)\left(\sqrt{2-\sqrt{3}}\right)}{\sqrt{2+\sqrt{3}}}\)

\(\frac{\left(2+\sqrt{3}\right)\left(\sqrt{4-2\sqrt{3}}\right)}{\sqrt{4+2\sqrt{3}}}\)

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

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

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

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