Bài làm:

\(\sqrt{14-8\sqrt{3}}-\sqrt{24-12\sqrt{3}}\)

\(=\sqrt{8-8\sqrt{3}+6}-\sqrt{18-12\sqrt{3}+6}\)

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

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

\(=-\sqrt{2}\)

Học tốt 

\(\sqrt{3.2-8\sqrt{3}+4.2}=\sqrt{\left(\sqrt{3}.\sqrt{2}\right)^2-2.2.\sqrt{2}.\sqrt{2}.\sqrt{3}+\left(2\sqrt{2}\right)^2}\)

\(=\sqrt{\left(2\sqrt{2}-\sqrt{6}\right)^2}-\sqrt{3^2.2-12\sqrt{3}+3.2}=2\sqrt{2}-\sqrt{6}-3\sqrt{2}+\sqrt{6}=-\sqrt{2}\)

Bài 2:

a: \(\frac{21+8\sqrt5}{4+\sqrt5}\cdot\sqrt{9-4\sqrt5}\)

\(=\frac{\left(4+\sqrt5\right)^2}{4+\sqrt5}\cdot\sqrt{\left(\sqrt5-2\right)^2}\)

\(=\left(4+\sqrt5\right)\left(\sqrt5-2\right)=4\sqrt5-8+5-2\sqrt5=2\sqrt5-3\)

b: \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)

\(=\sqrt{\left(\sqrt5-\sqrt3\right)^2}-\sqrt{\left(\sqrt5+\sqrt3\right)^2}\)

\(=\sqrt5-\sqrt3-\sqrt5-\sqrt3=-2\sqrt3\)

Bài 1:

a: \(\frac{\sqrt6-\sqrt{15}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt3\left(\sqrt2-\sqrt5\right)}{-\sqrt7\left(\sqrt2-\sqrt5\right)}=-\sqrt{\frac37}=-\frac{\sqrt{21}}{7}\)

b: \(\frac{10+2\sqrt{10}}{\sqrt5+\sqrt2}+\frac{8}{1-\sqrt5}\)

\(=\frac{2\sqrt5\left(\sqrt5+\sqrt2\right)}{\sqrt5+\sqrt2}-\frac{8\left(\sqrt5+1\right)}{\left(\sqrt5-1\right)\left(\sqrt5+1\right)}\)

\(=2\sqrt5-2\left(\sqrt5+1\right)=-2\)

c: \(\frac{\sqrt{3-\sqrt5}\left(3+\sqrt5\right)}{\sqrt{10}+\sqrt2}=\frac{\sqrt{6-2\sqrt5}\left(3+\sqrt5\right)}{\sqrt{20}+2}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}=\frac{\left(\sqrt5-1\right)\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}\)

\(=\frac{3\sqrt5+5-3-\sqrt5}{2\left(\sqrt5+1\right)}=\frac{2\sqrt5+2}{2\sqrt5+2}\)

=1

d: \(\frac{1}{\sqrt2+\sqrt{2+\sqrt3}}+\frac{1}{\sqrt2-\sqrt{2+\sqrt3}}\)

\(=\frac{\sqrt2}{2+\sqrt{4+2\sqrt3}}+\frac{\sqrt2}{2-\sqrt{4+2\sqrt3}}\)

\(=\frac{\sqrt2}{2+\sqrt{\left(\sqrt3+1\right)^2}}+\frac{\sqrt2}{2-\sqrt{\left(\sqrt3-1\right)^2}}=\frac{\sqrt2}{3+\sqrt3}+\frac{\sqrt2}{3-\sqrt3}\)

\(=\frac{\sqrt2\left(3-\sqrt3\right)+\sqrt2\left(3+\sqrt3\right)}{\left(3-\sqrt3\right)\left(3+\sqrt3\right)}=\frac{3\sqrt2-\sqrt6+3\sqrt2+\sqrt6}{9-3}=\frac{6\sqrt2}{6}=\sqrt2\)

Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)

\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=-\frac{3}{2}\)

a) Kết quả rút gọn xấu (+dài) nữa. (có thể đề sai)

b) 

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left[\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right)\)

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

c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)

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

a) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}=\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right].\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{1}{2}-2=-\frac{3}{2}\)

a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)

\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)

\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)

b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)

c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

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

a, =\(9\sqrt{2}\)

b, =21

a) \(=9\sqrt{2}\)

b) \(=21\)

học tốt.

e , \(\sqrt{11^2-\left(6\sqrt{2}\right)^2}\)

g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath

\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{\sqrt{2}}{2}\right)\cdot3\sqrt{6}\\ =36-36\sqrt{2}+30\sqrt{3}-3\sqrt{3}\\ =36-36\sqrt{2}+27\sqrt{3}\)