đề sai

+) \(\left(\sqrt{4}-\sqrt{3}\right)^2=4-2\sqrt{4\cdot3}+3=7-2\sqrt{7}=\sqrt{49}-\sqrt{48}\)

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

\(=4\sqrt{2}-6\sqrt{6}+9-4\sqrt{2}+6\sqrt{6}\)

\(=9\)

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

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

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

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

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

Sửa đề: Chứng minh \(\left(\sqrt{7+4\sqrt{3}}+\sqrt{8-2\sqrt{15}}\right)-\left(\sqrt{8+2\sqrt{15}}-\sqrt{7-4\sqrt{3}}\right)=\left(\sqrt{3}-1\right)^2\)

Ta có: \(VT=\left(\sqrt{7+4\sqrt{3}}+\sqrt{8-2\sqrt{15}}\right)-\left(\sqrt{8+2\sqrt{15}}-\sqrt{7-4\sqrt{3}}\right)\)

\(=\left(\sqrt{4+2\cdot2\cdot\sqrt{3}+3}+\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\right)-\left(\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\right)\)

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

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

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

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

\(=4-2\sqrt{3}\)

\(=3-2\cdot\sqrt{3}\cdot1+1\)

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

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

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

\(=2\sqrt{3}\)

4) \(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)

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

5) \(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}\)

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

2: \(\dfrac{\sqrt{108}}{\sqrt{3}}=6\)

13: \(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)

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

\(=-\sqrt{5}\)

14: \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

=2

12.

\(\dfrac{\sqrt{108}}{\sqrt{3}}=\dfrac{\sqrt{36}.\sqrt{3}}{\sqrt{3}}=\sqrt{36}=6\)

13.

\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}\)

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

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

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

\(=-\sqrt{5}\)

14.

\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\)

\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{16-15}\)

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

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

\(=2\)

Bài 1

a) Đặt VT = A

<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)

<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)

<=> 2A = \(\left(5-3\right)^2=4\)

<=> A = 2

b) Đặt VT = B

<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)

<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)

<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)

<=> B = 8 

Bài 2

Đặt VT = A

<=> A² = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)

<=> A² = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)

<=> \(A=\sqrt{\sqrt{5}+1}\)

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\\ =\sqrt{\sqrt{5^2}+2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}-\sqrt{\sqrt{5^2}-2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\left|\sqrt{5}+\sqrt{3}\right|-\left|\sqrt{5}-\sqrt{3}\right|\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}\\ =2\sqrt{3}\)

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

a) \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)

\(=\sqrt{5-2\cdot\sqrt{5\cdot3}+3}-\sqrt{5+2\cdot\sqrt{5\cdot3}+1}\)

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

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

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

b. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

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

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

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

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

\(=2\sqrt{2}\)

Bạn xem lại đề. Biểu thức trong căn thứ 2 âm nên biểu thức B không tồn tại. Có phải số 8 bạn nên sửa thành 9?

\(a,\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{3}+\sqrt{5}-\left(\sqrt{5}+1\right)=\sqrt{3}-1\\ b,=3-2\sqrt{2}-\left(3\sqrt{2}+1\right)=2-5\sqrt{2}\\ c,=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\\ d,=\sqrt{11}+1-\left(\sqrt{11}-1\right)=2\\ e,=\sqrt{7}-\sqrt{3}-\left(\sqrt{7}-\sqrt{2}\right)=\sqrt{2}-\sqrt{3}\)