a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)

\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)

\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)

b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)

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

\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)

b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)

c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)

\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)

\(=21-14\sqrt{2}+14\sqrt{2}\)

=21

\(=\left(3\sqrt{2}-2\sqrt{2}+\sqrt{14}\right).\sqrt{2}-\sqrt{7}\\ =\left(\sqrt{2}+\sqrt{14}\right).\sqrt{2}-\sqrt{7}\\ =2+2\sqrt{7}-\sqrt{7}\\ =2+\sqrt{7}\)

\(\sqrt{8-2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)

\(\sqrt{29-12\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)

\(=\left|2\sqrt{5}-3\right|-\left|\sqrt{5}-2\right|=2\sqrt{5}-3-\sqrt{5}+2=\sqrt{5}-1\)

b)\(=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c)\(=\sqrt{\left(2\sqrt{5}-3\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(=2\sqrt{5}-3-\sqrt{5}+2\)
\(=\sqrt{5}-1\)

b,\(\sqrt{8-2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)

c,\(\sqrt{29-12\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{\left(3-2\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}=3-2\sqrt{5}-\sqrt{5}+2=5-3\sqrt{5}\)

Bạn ghi lại đề dùm

\(\sqrt{8-2\sqrt{7}-\left[\left(\sqrt{7}+1\right)^2\right]}\)

\(\sqrt{8-2\sqrt{7}-\sqrt{7}-1}\)

\(\Leftrightarrow\sqrt{7-\sqrt{7}}\)

\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

= \(\sqrt{7}-1-\left(1+\sqrt{7}\right)\)

\(=\sqrt{7}-1-\sqrt{7}-1\)

= -2

\(\sqrt{8-2\sqrt{7}}-\sqrt{23-8\sqrt{7}}=\) \(\sqrt{1-2\sqrt{7}+7}-\sqrt{7-2.4.\sqrt{7}+16}\)

\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-4\right)^2}\)

\(=\sqrt{7}-1-\left(-\sqrt{7}+4\right)\)

\(=\sqrt{7}-1+\sqrt{7}-4\)\(=2\sqrt{7}-5\)

chúc bn học tốt

=\(\sqrt{\left(\sqrt{7}-1\right)^2}\)- \(\sqrt{\left(4-\sqrt{7}\right)^2}\)

= \(\sqrt{7}\)- 1 - 4 + \(\sqrt{7}\)

= \(2\sqrt{7}\)-5

đ/á ra hơi kì

#mã mã#

mình nghĩ là hai bn làm đúng đó

A= \(\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\)\(1+\sqrt{7}+\sqrt{7}-1=2\sqrt{7}\)

\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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