a: \(=\dfrac{\sqrt{2}\left(2\sqrt{2}+3\right)+2\sqrt{2}-3}{-1}\)

\(=\dfrac{4+3\sqrt{2}+2\sqrt{2}-3}{-1}=-1-5\sqrt{2}\)

b: \(=\dfrac{1}{\sqrt{10}+\sqrt{6}}-\dfrac{1}{\sqrt{10}-\sqrt{6}}\)

\(=\dfrac{\sqrt{10}-\sqrt{6}-\sqrt{10}-\sqrt{6}}{4}=\dfrac{-2\sqrt{6}}{4}=-\dfrac{\sqrt{6}}{2}\)

c: \(\dfrac{-2}{3\sqrt{8}}+\dfrac{1}{3-2\sqrt{2}}\)

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

\(=\dfrac{10\sqrt{2}-6}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{10-3\sqrt{2}}{6\left(3-2\sqrt{2}\right)}=\dfrac{18+11\sqrt{2}}{6}\)

1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)

2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)

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

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

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

\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)

3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)

\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)

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

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

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

\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)

3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

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

=-2

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

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

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

=4

a: Ta có: \(A=\sqrt{8}-2\sqrt{18}+3\sqrt{50}\)

\(=2\sqrt{2}-6\sqrt{2}+15\sqrt{2}\)

\(=11\sqrt{2}\)

b: Ta có: \(B=\sqrt{125}-10\sqrt{\dfrac{1}{20}}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)

\(=5\sqrt{5}-\sqrt{5}+\sqrt{5}-1\)

\(=5\sqrt{5}-1\)

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

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

=4

a: \(=12\sqrt{80}=48\sqrt{5}\)

b: \(=2\sqrt{5}\cdot2\sqrt{3}-10=4\sqrt{15}-10\)

c: =20-9=11

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

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

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

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

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

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

a: \(\frac{\sqrt{3-\sqrt5}\cdot\left(3+\sqrt5\right)}{\sqrt{10}+\sqrt2}\)

\(=\frac{\sqrt{6-2\sqrt5}\cdot\left(3+\sqrt5\right)}{\sqrt{20}+\sqrt4}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+\sqrt1\right)}\)

\(=\frac{\left(\sqrt5-1\right)^{}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+\sqrt1\right)}=\frac{3\sqrt5+5-3-\sqrt5}{2\left(\sqrt5+1\right)}=\frac{2\sqrt5+2}{2\sqrt5+2}=1\)

b: \(\sqrt{8\sqrt3}-\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)

\(=2\sqrt{2\sqrt3}-5\sqrt{2\sqrt3}+4\sqrt{\sqrt{64}\cdot\sqrt3}\)

\(=-3\sqrt{2\sqrt3}+4\sqrt{8\sqrt3}\)

\(=-3\sqrt{2\sqrt3}+4\cdot2\sqrt{2\sqrt3}=5\sqrt{2\sqrt3}\)

c: \(\sqrt{2-\sqrt3}\cdot\left(\sqrt5+\sqrt2\right)\)

\(=\frac{\sqrt{4-2\sqrt3}\cdot\left(\sqrt5+\sqrt2\right)}{\sqrt2}=\frac{\left(\sqrt3-1\right)\left(\sqrt5+\sqrt2\right)}{\sqrt2}=\frac{\left(\sqrt6-\sqrt2\right)\left(\sqrt5+\sqrt2\right)}{2}\)

\(=\frac{\sqrt{30}+2\sqrt3-\sqrt{10}-2}{2}\)

d: \(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\)

\(=\frac{\sqrt{6-2\sqrt5}+\sqrt{6+2\sqrt5}}{\sqrt2}=\frac{\sqrt{\left(\sqrt5-1\right)^2}+\sqrt{\left(\sqrt5+1\right)^2}}{\sqrt2}\)

\(=\frac{\sqrt5-1+\sqrt5+1}{\sqrt2}=\frac{2\sqrt5}{\sqrt2}=\sqrt{10}\)

e: Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)

=>\(A^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{16-\left(10+2\sqrt5\right)}\)

=>\(A^2=8+2\cdot\sqrt{6-2\sqrt5}\)

=>\(A^2=8+2\cdot\sqrt{\left(\sqrt5-1\right)^2}=8+2\left(\sqrt5-1\right)=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

=>\(A=\sqrt5+1\)

f: \(\left(5+2\sqrt6\right)\left(49-20\sqrt6\right)\cdot\sqrt{5-2\sqrt6}\)

\(=\left(245-100\sqrt6+98\sqrt6-240\right)\cdot\sqrt{\left(\sqrt3-\sqrt2\right)^2}\)

\(=\left(5-2\sqrt6\right)\left(\sqrt3-\sqrt2\right)=\left(\sqrt3-\sqrt2\right)^2\)

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

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

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

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

\(=\frac{\sqrt2}{2+\sqrt3+1^{}}+\frac{\sqrt2}{2-\sqrt3+1}=\frac{\sqrt2}{3+\sqrt3}+\frac{\sqrt2}{3-\sqrt3}=\frac{\sqrt2\left(3-\sqrt3\right)+\sqrt2\left(3+\sqrt3\right)}{9-3}\)

\(=\frac{3\sqrt2-\sqrt6+3\sqrt2+\sqrt6}{6}=\frac{6\sqrt2}{6}=\sqrt2\)

i: \(\frac{\left(\sqrt5+2\right)^2-8\sqrt5}{2\sqrt5-4}\)

\(=\frac{9+4\sqrt5-8\sqrt5}{2\left(\sqrt5-2\right)}\)

\(=\frac{9-4\sqrt5}{2\left(\sqrt5-2\right)}=\frac{\left(\sqrt5-2\right)^2}{2\left(\sqrt5-2\right)}=\frac{\sqrt5-2}{2}\)

k: \(\sqrt{14-8\sqrt3}-\sqrt{24-12\sqrt3}\)
\(=\sqrt{8-2\cdot2\sqrt2\cdot\sqrt6+6}-\sqrt{6\left(4-2\sqrt3\right)}\)
\(=\sqrt{\left(2\sqrt2-\sqrt6\right)^2}-\sqrt6\left(\sqrt3-1\right)=2\sqrt2-\sqrt6-\sqrt{18}+\sqrt6=2\sqrt2-3\sqrt2=-\sqrt2\)

l: \(\frac{4}{\sqrt3+1}+\frac{1}{\sqrt3-2}+\frac{6}{\sqrt3-3}\)

\(=\frac{4\left(\sqrt3-1\right)}{3-1}-\frac{1\left(2+\sqrt3\right)}{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}-\frac{6\left(3+\sqrt3\right)}{9-3}\)

\(=2\left(\sqrt3-1\right)-\left(2+\sqrt3\right)-\left(3+\sqrt3\right)=2\sqrt3-2-2-\sqrt3-3-\sqrt3\)

=-7

m: \(\left(\sqrt2+1\right)^3-\left(\sqrt2-1\right)^3\)

\(=\left(2\sqrt2+3\cdot2\cdot1+3\cdot\sqrt2\cdot1+1\right)-\left(2\sqrt2-3\cdot2\cdot1+3\cdot\sqrt2\cdot1-1\right)\)

\(=\left(5\sqrt2+7\right)-\left(5\sqrt2-7\right)=14\)