a) \(\sqrt{\left(4-\sqrt{15}\right)^{2^{ }}}+\sqrt{15}\)

=\(\left|4-\sqrt{15}\right|+\sqrt{15}\)

= \(4-\sqrt{15}+\sqrt{15}\) ( vì 4 =\(\sqrt{16}\) mà \(\sqrt{16}>\sqrt{15}\) )

=4

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

=\(\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\) ( vì \(1< \sqrt{3}< 2\))

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

=1

a) Ta có: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)

\(=\left|4-\sqrt{15}\right|+\sqrt{15}\)

\(=4-\sqrt{15}+\sqrt{15}\)

=4

b) Ta có: \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\)

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

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

\(=1\)

(B)

x 6cm 4cm Theo định lý Py - ta - go :

x² = 4² + 6²

⇒ x² = 16 + 36

⇒ x² = 52

⇒ x =

Ta có 

\(\left(\sqrt{25-x^2}-\sqrt{15-x^2}\right)\left(\sqrt{25-x^2}+\sqrt{15-x^2}\right)=25-x^2-15+x^2=10\)

\(\Rightarrow\sqrt{25-x^2}+\sqrt{15-x^2}=5\)

Ta có:

\(\left(\sqrt{25-x^2}-\sqrt{15-x^2}\right)\left(\sqrt{25-x^2}+\sqrt{15-x^2}\right)=25-x^2-\left(15-x^2\right)=10\)

\(\Rightarrow y=\sqrt{25-x^2}+\sqrt{15-x^2}=\dfrac{10}{2}=5\)

ko biết

a) \(\sqrt{5+\sqrt{21}}-\sqrt{6-\sqrt{35}}\) = \(\dfrac{\sqrt{10+2\sqrt{21}}}{\sqrt{2}}-\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}\)

= \(\dfrac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}}{\sqrt{2}}\)

= \(\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{2}}\) = \(\dfrac{\sqrt{7}+\sqrt{3}-\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{2}}\)

= \(\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\)

câu b) hình như đề sai