a) \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)

\(=\sqrt{14}\cdot\sqrt{5-\sqrt{21}}+\sqrt{6}\cdot\sqrt{5-\sqrt{21}}\)

\(=\sqrt{14\cdot\left(5-\sqrt{21}\right)}+\sqrt{6\cdot\left(5-\sqrt{21}\right)}\)

\(=\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\)

\(=\sqrt{7^2-2\cdot7\cdot\sqrt{21}+\left(\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}\right)^2-2\cdot3\cdot\sqrt{21}+3^2}\)

\(=\sqrt{\left(7-\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}-3\right)^2}\)

\(=\left|7-\sqrt{21}\right|+\left|\sqrt{21}-3\right|\)

\(=7-\sqrt{21}+\sqrt{21}-3\)

\(=4\)

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

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

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

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

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

\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)

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

\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)

\(=5-\sqrt{15}+\sqrt{15}-3\)

\(=2\)

Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)

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

=1

a,A.√2= √(4+2√3)-√(4-2√3)

= √(1+√3)² -√( √3 -1)²

= 1+√3-√3+1= 2 

=> A= 2/√2=√2

B²= (4+√15)².(4-√15).(√10-√6)²

= (4+√15).1.(16-4√15)

= (4+√15).(4-√15).4

= 4

=> B = √4 = 2

\(6\sqrt{\frac{3}{4}}+10\sqrt{\frac{12}{25}}-15\sqrt{\frac{16}{3}}+9\sqrt{\frac{4}{3}}\)

\(=6\cdot\frac{\sqrt{3}}{2}+10\cdot\frac{2\sqrt{3}}{5}-15\cdot\frac{4}{\sqrt{3}}+9\cdot\frac{2}{\sqrt{3}}\)

\(=3\sqrt{3}+4\sqrt{3}-20\sqrt{3}+6\sqrt{3}=-7\sqrt{3}\)

Trả lời:

\(6\sqrt{\frac{3}{4}}+10\sqrt{\frac{12}{25}}-15\sqrt{\frac{16}{3}}+9\sqrt{\frac{4}{3}}\)

\(=6.\frac{\sqrt{3}}{\sqrt{4}}+10.\frac{\sqrt{12}}{\sqrt{25}}-15.\frac{\sqrt{16}}{\sqrt{3}}+9.\frac{\sqrt{4}}{\sqrt{3}}\)

\(=6.\frac{\sqrt{3}}{2}+10.\frac{\sqrt{2^2.3}}{5}-15.\frac{4}{\sqrt{3}}+9.\frac{2}{\sqrt{3}}\)

\(=3\sqrt{3}+10.\frac{2\sqrt{3}}{5}-15.\frac{4\sqrt{3}}{3}+9.\frac{2\sqrt{3}}{3}\)

\(=3\sqrt{3}+4\sqrt{3}-20\sqrt{3}+6\sqrt{3}\)

\(=\left(3+4-20+6\right).\sqrt{3}=-7\sqrt{3}\)

9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)

10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)

12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)

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

=9-3=6

13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)

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

a) Đặt A=\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

<=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)

= \(\sqrt{7}+1-\sqrt{7}+1=2\)

=> \(A=\frac{2}{\sqrt{2}}\sqrt{2}\)

b) Ta đặt \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

=> \(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)

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

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

=>  B=\(\sqrt{5}+1\)

c) Ta xét \(A=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\)

=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{3}\cdot\sqrt{5}}+\sqrt{8-2\sqrt{3}\cdot\sqrt{5}}\)

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

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

=> A=\(\sqrt{5}\)

Ta có : \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

= \(A-\sqrt{6-2\sqrt{5}}\)

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

Phần a) chỗ cuối viết thiếu dấu =.

Sẽ là A=\(\sqrt{2}\)nha

cám ơn bạn nha

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

\(b,=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1}+\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}+1+\sqrt{3}-1\\ =2\sqrt{3}\)

\(c,=x-4+\sqrt{\left(4^2-2.4.x+x^2\right)}\\ =x-4+\sqrt{\left(4-x\right)^2}\\ =x-4+\left|4-x\right|\\ =x-4+x-4=2x-8\)    (vì \(x>4\) )

@seven