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

b) \(=\dfrac{\sqrt{5}+2}{5-4}-\dfrac{\sqrt{5}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{5}+2-\sqrt{5}=2\)

\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\)    ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))

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

\(=\sqrt{4\cdot\sqrt{7}}\)

\(2,\sqrt{\left(3\sqrt{5}-5\sqrt{2}\right)^2}-\sqrt{\left(5\sqrt{2}+3\sqrt{5}\right)^2}\)

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

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

\(\Leftrightarrow\left(3\sqrt{5}-5\sqrt{2}\right)^2-\left(5\sqrt{2}+3\sqrt{5}\right)^2\)

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

\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)

\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)

\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)

\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)

\(\Leftrightarrow4\sqrt{21}\)

cuối lười tính nên thôi nhá :>

tks :>

\(a.\sqrt{4-\sqrt{15}}.\sqrt{4+\sqrt{15}}\)

\(=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}=\sqrt{\left(16-15\right)}=\sqrt{1}=1\)

\(b.\sqrt{7-\sqrt{47}}.\sqrt{14+2\sqrt{47}}\)

\(=\sqrt{7-\sqrt{47}}.\sqrt{2\left(7-\sqrt{47}\right)}\)

\(=\sqrt{2\left(7-\sqrt{47}\right)\left(7+\sqrt{47}\right)}=\sqrt{2\left(49-47\right)}=\sqrt{2^2}=\sqrt{4}=2\)

\(c.\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(=\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)

\(=\sqrt{16-\left(\sqrt{10+2\sqrt{5}}\right)^2}\)

\(=\sqrt{16-10-2\sqrt{5}}=\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)

\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{48-20-10\sqrt{3}}}}\)

\(=\sqrt{5\sqrt{3}+\sqrt{5\sqrt{28-10\sqrt{3}}}}\)

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

\(=\sqrt{5\sqrt{3}+\sqrt{5\left(5-\sqrt{3}\right)}}=\sqrt{5\sqrt{3}+\sqrt{25-5\sqrt{3}}}\)

Trần Đức Thắng lm nốt đi

\(a)\)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)

\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)

\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)

\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)

\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)

\(=\)\(3\sqrt{6}+9\)

Chúc bạn học tốt ~ 

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

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

\(=\)\(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\) ( vì \(2=\sqrt{4}>\sqrt{3}\) ) 

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

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

\(=\)\(1\)

Chúc bạn học tốt ~ 

PS : mới lớp 8 sai thì thông cảm >.< 

\(\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}\) + \(\sqrt{\left(\sqrt{24}-\sqrt{9}\right)^2}\)

\(\sqrt{9}-\sqrt{6}\) + \(\sqrt{24}-\sqrt{9}\) vì  \(\sqrt{9}>\sqrt{6}\),\(\sqrt{24}>\sqrt{9}\)

\(3-\sqrt{6}\) + \(2\sqrt{6}-3\)

= \(\sqrt{6}\)

a/ Đặt \(\hept{\begin{cases}\sqrt{3+\sqrt{5}}=a\\\sqrt{3-\sqrt{5}}=b\end{cases}}\)

Khi đó ta có a² + b² = 6; ab = 2; a + b = \(\sqrt{10}\) ; a - b = \(\sqrt{2}\); a² - b² = \(2\sqrt{5}\)

Ta có cái ban đầu

\(=\frac{a^2}{\sqrt{10}+a}-\frac{b^2}{\sqrt{10}+b}\)=

\(\frac{\sqrt{10}a^2+a^2b-\sqrt{10}b^2-ab^2}{10+\sqrt{10}a+\sqrt{10}b+ab}\)

\(=\frac{10\sqrt{2}+2\sqrt{2}}{10+10+2}=\frac{6\sqrt{2}}{11}\)

Câu còn lại làm tương tự

Vd1: 

d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)

\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)

\(\Leftrightarrow x=6\)