~~~~~a)~~~~~

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

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

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

\(=2.\sqrt{\frac{1}{2}}=\sqrt{2}\)

*****b)*****

(Hình như đề có cái gì đó sai sai hả bạn?)

~~~~~c)~~~~~

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

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

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

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

\(=-2\)

*****d)*****

     \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

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

\(=2\sqrt{2}-\sqrt{5}-2\sqrt{2}-3\sqrt{5}\)

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

(Chúc bạn học tốt và tíck cho mìk vs nhé ~~~~~bạn xem lại câu b hộ mình luôn nha~~~~~!)

Bài 1:

a) Ta có: \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

\(=\sqrt{45-2\cdot\sqrt{45}\cdot1+1}-\sqrt{9-2\cdot\sqrt{9}\cdot\sqrt{20}+20}\)

\(=\sqrt{\left(\sqrt{45}-1\right)^2}-\sqrt{\left(3-\sqrt{20}\right)^2}\)

\(=\left|\sqrt{45}-1\right|-\left|3-\sqrt{20}\right|\)

\(=\sqrt{45}-1-3+\sqrt{20}\)

\(=\sqrt{45}+\sqrt{20}-4\)

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

b) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{8}+8}-\sqrt{45+2\cdot\sqrt{45}\cdot\sqrt{8}+8}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{8}\right)^2}-\sqrt{\left(\sqrt{45}+\sqrt{8}\right)^2}\)

\(=\left|\sqrt{5}-\sqrt{8}\right|-\left|\sqrt{45}+\sqrt{8}\right|\)

\(=\sqrt{8}-\sqrt{5}-\sqrt{45}-\sqrt{8}\)

\(=-\sqrt{5}-\sqrt{45}=-\sqrt{5}\left(1+\sqrt{9}\right)=-4\sqrt{5}\)

c) Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{7+4\sqrt{3}}\)

\(=\left(3-\sqrt{2}\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\)

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

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

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

d) Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10+2\sqrt{21}}\)

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

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

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

\(=\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2=7-3=4\)

à

a/\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)

b/ \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)

\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)

\(=2\sqrt{3}\cdot6\sqrt{3}=2\cdot6\cdot3=36\)

c/ \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)

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

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

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

d/ \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{8-4\sqrt{10}+5}-\sqrt{45+12\sqrt{10}+8}\)

\(=\sqrt{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2\cdot5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\cdot2\sqrt{5\cdot2}+\left(2\sqrt{2}\right)^2}\)

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

\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}\)

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

#)Giải :

 \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)

Mình làm luôn nhé :

\(\sqrt{45-2.3\sqrt{5}+1}-\sqrt{20-2.3.2\sqrt{5}+9}\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5-\sqrt{45+2.2.\sqrt{2}.3\sqrt{5}+8}}\left(\sqrt{3}+\sqrt{5}\right).\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{7+2.\sqrt{7}.\sqrt{3}+3}\) Tới đây dễ rồi , bạn tự nhóm HĐT là ra ::v

a) \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-2\sqrt{40}}-\sqrt{53+12\sqrt{10}}\)

\(=\sqrt{\left(\sqrt{8}\right)^2-2.\sqrt{8}.\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2.3\sqrt{5}.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{8}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)

\(=\left|\sqrt{8}-\sqrt{5}\right|-\left|3\sqrt{5}+2\sqrt{2}\right|\)

= √8 - √5 - 3√5 - 2√2 = -4√5

b) (1+√3-√2).(1+√3+√2)= [(1+√3)^2-(√2)^2] = 4+2√3-2=2+2√3

a) =sprt{13-=sprt{160}} - =sprt{53+4=sprt{90}}

= =sprt{(=sprt{8} - =sprt{5})² } - =sprt{(=sprt{45} + =sprt{8})² }

= =sprt{8} - =sprt{5} - =sprt{45} - =sprt{8}

= -3=sprt{5}

b) ( 1 + =sprt{3} - =sprt{2} )( 1+ =sprt{3} + =sprt{2} )

=  ( 1 + =sprt{3} )² - (=sprt{2})²

= 4 + 2=sprt{3} - 2

=2 + 2=sprt{3}