\(\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)\left(\sqrt{\sqrt{5}-1}\right)}{\left(\sqrt{\sqrt{5}+1}\right)\left(\sqrt{\sqrt{5}-1}\right)}-\sqrt{3-2\sqrt{2}}\)
\(=\dfrac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{\sqrt{5-1}}-\sqrt{2-2\sqrt{2}+1}\)
Đặt: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}\)
\(\Rightarrow A^2=3+\sqrt{5}+2\sqrt{\left(3+\sqrt{5}\right)\left(7-3\sqrt{5}\right)}+7-3\sqrt{5}\)
\(A^2=10-2\sqrt{5}+2\sqrt{6-2\sqrt{5}}\)
\(A^2=10-2\sqrt{5}+2\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1}\)
\(A^2=10-2\sqrt{5}+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(A^2=10-2\sqrt{5}+2\left|\sqrt{5}-1\right|\)
\(A^2=10-2\sqrt{5}+2\sqrt{5}-2\)
\(A^2=8\Rightarrow A=2\sqrt{2}\)
Thay vào ta có:
\(\dfrac{2\sqrt{2}}{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}-\left|\sqrt{2}-1\right|\)
\(=\sqrt{2}-\left(\sqrt{2}-1\right)=\sqrt{2}-\sqrt{2}+1=1\)
√√5+2+√√5−2√√5+1−√3−2√2
=(√√5+2+√√5−2)(√√5−1)(√√5+1)(√√5−1)−√3−2√2
=√3+√5+√7−3√5√5−1−√2−2√2+1
Đặt: A=√3+√5+√7−3√5
⇒A2=3+√5+2√(3+√5)(7−3√5)+7−3√5
A2=10−2√5+2√6−2√5
A2=10−2√5+2√(√5)2−2√5+1
A2=10−2√5+2√(√5−1)2
A2=10−2√5+2|√5−1|
A2=10−2√5+2√5−2
A2=8⇒A=2√2
Thay vào ta có:
2√22−√(√2−1)2
=√2−|√2−1|
=√2−(√2−1)=√2−√2+1=1
