a)

\((\sqrt{3}-2\sqrt{12}+2\sqrt{4})(\sqrt{27}+\sqrt{144}-2\sqrt{16})\)

\(=(\sqrt{3}-4\sqrt{3}+4)(3\sqrt{3}+12-8)\)

\(=(-3\sqrt{3}+4)(3\sqrt{3}+4)=4^2-(3\sqrt{3})^2=16-27=-11\)

b)

\((2\sqrt{5}+2\sqrt{3})^2-4\sqrt{60}\)

\(=(2\sqrt{5})^2+2.2\sqrt{5}.2\sqrt{3}+(2\sqrt{3})^2-8\sqrt{15}\)

\(=32+8\sqrt{15}-8\sqrt{15}=32\)

c)

\(\sqrt{6}(3\sqrt{12}-4\sqrt{3}+\sqrt{48}-5\sqrt{6})\)

\(=3\sqrt{72}-4\sqrt{18}+\sqrt{6.48}-5.\sqrt{36}\)

\(=18\sqrt{2}-12\sqrt{2}+12\sqrt{2}-30=18\sqrt{2}-30\)

d)

\((\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})\)

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

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

e) Biểu thức bên trong căn lớn âm nên biểu căn bậc 2 không có nghĩa

f)

\((\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}).\frac{1}{\sqrt{3}+5}\)

\(=(\frac{2\sqrt{3}+15}{3-\sqrt{3}}+\frac{3}{\sqrt{3}-2}).\frac{1}{\sqrt{3}+5}\)

\(=\frac{2\sqrt{3}+15)(\sqrt{3}-2)+3(3-\sqrt{3})}{(3-\sqrt{3})(\sqrt{3}-2)}.\frac{1}{\sqrt{3}+5}\)

\(=\frac{-15+8\sqrt{3}}{(-9+5\sqrt{3})(\sqrt{3}+5)}=\frac{-15+8\sqrt{3}}{-30+16\sqrt{3}}=\frac{-15+8\sqrt{3}}{2(-15+8\sqrt{3})}=\frac{1}{2}\)

\(\sqrt{7+2\sqrt{3}}-\sqrt{7-2\sqrt{3}};\left(\sqrt{7+\sqrt{12}}-\sqrt{7-\sqrt{12}}\right)^2=7+\sqrt{12}-\sqrt{12}+7-2\sqrt{\left(7+\sqrt{12}\right)\left(7-\sqrt{12}\right)}=14-2\sqrt{37}\Rightarrow\sqrt{7+\sqrt{12}}-\sqrt{7-\sqrt{12}}=\sqrt{14-2\sqrt{37}}\)

a) \(\sqrt{10+2\sqrt{14}}\cdot\sqrt{10+2\sqrt{14}}\)

\(=\sqrt{\left(10+2\sqrt{14}\right)^2}\)

\(=10+2\sqrt{14}\)

b) \(\sqrt{7+\sqrt{12}}-\sqrt{7-\sqrt{12}}\)

\(=\sqrt{\left(\sqrt{7+\sqrt{12}}-\sqrt{7-\sqrt{12}}\right)^2}\)

\(=\sqrt{7+\sqrt{12}+7-\sqrt{12}-2\sqrt{\left(7+\sqrt{12}\right)\left(7-\sqrt{12}\right)}}\)

\(=\sqrt{14-2\sqrt{49-12}}\)

\(=\sqrt{14-2\sqrt{37}}\)

\(\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\)

\(=\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\)

\(=\dfrac{\sqrt{5}}{\sqrt{4}}\)

\(=\dfrac{\sqrt{5}}{2}\)

\(\dfrac{\sqrt{6}-\sqrt{15}}{\sqrt{35}-\sqrt{14}}\)

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

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

\(=-\dfrac{\sqrt{3}}{\sqrt{7}}\)

\(=-\dfrac{\sqrt{21}}{7}\)

____________

\(\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\)

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

\(=\dfrac{\sqrt{5}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{10}}{2}\)

B=\(\frac{3\sqrt{x}+4}{3\sqrt{x}-2}-\frac{42\sqrt{x}+34}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{(3\sqrt{x}+4)(5\sqrt{x}+7)-42\sqrt{x}-34}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{15x+20\sqrt{x}+21\sqrt{x}+28-42\sqrt{x}-34}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{15x-\sqrt{x}-6}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+3\right)}{\left(3\sqrt{x}-2\right)\left(5\sqrt{x}+7\right)}=\frac{5\sqrt{x}+3}{5\sqrt{x}+7}\)

\(a,\dfrac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}\\ =\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{\sqrt{2}.\sqrt{4}-\sqrt{3}.\sqrt{4}}\\ =\dfrac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2^2}}\\ =\dfrac{\sqrt{5}}{2}\)

\(b,\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\\ =\dfrac{\sqrt{5}.\sqrt{3}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}\)

\(c,\dfrac{5+\sqrt{5}}{\sqrt{10}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}=\dfrac{\sqrt{5}}{\sqrt{2}}\)

\(a,=\dfrac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}\\ =\dfrac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}}{2}\)

\(b,=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\\ =\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}\\ =\dfrac{\sqrt{3}}{\sqrt{7}}=\dfrac{\sqrt{21}}{7}\)

\(c,=\dfrac{\sqrt{5}.\sqrt{5}+\sqrt{5}}{\sqrt{2}.\sqrt{5}+\sqrt{2}}\\ =\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{2}\left(\sqrt{5}+1\right)}\\ =\dfrac{\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{10}}{2}\)

1.

a, \(2\sqrt{18}-4\sqrt{50}-3\sqrt{32}=6\sqrt{2}-20\sqrt{2}-12\sqrt{2}=-2\sqrt{2}\)

b, \(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}+3\right)^2}\)

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

\(=-\sqrt{5}+3+\sqrt{5}+3=6\)

c, \(\dfrac{\sqrt{10}+10}{1+\sqrt{10}}-\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}=\dfrac{\sqrt{10}\left(1+\sqrt{10}\right)}{1+\sqrt{10}}-\dfrac{\sqrt{10}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{5}-\sqrt{2}}\)

\(=\sqrt{10}-\sqrt{10}=0\)

2.

ĐK: \(x\in R\)

\(\sqrt{9x^2-30x+25}=5\)

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

\(\Leftrightarrow\left|3x-5\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=5\\3x-5=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=0\end{matrix}\right.\)

Vậy ...

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

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

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

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

6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)

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