Question

Tính giá trị biểu thức:

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

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

Answer 1

Lời giải:
\(\sqrt{2}A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}\)

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

$\Rightarrow A\geq \sqrt{2}$

\(B=2\sqrt{6}-4\sqrt{2}+(9+4\sqrt{2})-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}\)

\(=9\)

Answer 2

a)ta có:\(A^2=\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\)=\(2+\sqrt{3}+2-\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)

=\(4-2\sqrt{1}=4-2=2\)

\(\Rightarrow A=\pm\sqrt{2}\) mà A>0\(\Rightarrow A=\sqrt{2}\)

b)B=\(2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)=9