b) Ta có: \(B=\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

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

\(=2\sqrt{7}\)

d) Ta có: \(D=\sqrt{x^2-6x+9}-x\)

\(=\left|x-3\right|-x\)

\(=\left[{}\begin{matrix}x-3-x=-3\left(x\ge3\right)\\3-x-x=-2x+3\left(x< 3\right)\end{matrix}\right.\)

\(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

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

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

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

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

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

\(=\sqrt{7}+\sqrt{7}=2\sqrt{7}\)

Ta có: \(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

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

\(=2\sqrt{7}\)

đặt \(A=\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)

\(=>A^2=10-2\sqrt{21}+10+2\sqrt{21}+2\sqrt{\left(10-2\sqrt{21}\right)\left(10+2\sqrt{21}\right)}\)

\(=>A^2=20+2\sqrt{10^2-\left(2\sqrt{21}\right)^2}=20+2\sqrt{16}=20+2.4=28\)

\(=>A=\sqrt{28}=2\sqrt{7}\)

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

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

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

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

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

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

\(b,\)

\(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}=\left[\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\right].\sqrt{10-2\sqrt{21}}\)

\(=\sqrt{7}\left(\sqrt{7}+\sqrt{3}\right)\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7.3}+\left(\sqrt{3}\right)^2}\)

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

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

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

a) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)

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

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

=3-1=2

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

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

\(=4\sqrt{7}\)

\(B=10+5\sqrt{3}+10-5\sqrt{3}=20\)

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

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

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

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

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

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

cái x trục căn thức trong căn đi rồi thay vô A

giải chi tiết ra jùm cái

\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\)    ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))

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

\(=\sqrt{4\cdot\sqrt{7}}\)

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

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

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

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

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

\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)

\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)

\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)

\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)

\(\Leftrightarrow4\sqrt{21}\)

cuối lười tính nên thôi nhá :>

tks :>

a, \(\sqrt{25}-3\sqrt{\dfrac{4}{9}}=5-3.\dfrac{2}{3}=3\)

b, \(\left(2-\dfrac{5}{3}\right):\left(\dfrac{2}{7}+\dfrac{5}{21}-1\right)\)

\(=\dfrac{1}{3}:\dfrac{6+5-21}{21}\)

\(=-\dfrac{1}{3}.\dfrac{21}{10}\)

\(=-\dfrac{7}{10}\)