Question

Answer 1

45:

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

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

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

=1

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

\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{2}\)

=-1

Answer 2

45.

a, \(\left[\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right]:\sqrt{2}\)

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

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

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

Answer 3

45.

c, \(\left[\sqrt{8-\sqrt{15}}+\sqrt{8+\sqrt{15}}\right]:\sqrt{5}\)

\(=\dfrac{\sqrt{16-2\sqrt{15}}+\sqrt{16+2\sqrt{15}}}{\sqrt{10}}\)

\(=\dfrac{\sqrt{\left(\sqrt{15}-1\right)^2}+\sqrt{\left(\sqrt{15}+1\right)^2}}{\sqrt{10}}\)

\(=\dfrac{\sqrt{15}-1+\sqrt{15}+1}{\sqrt{10}}\)

\(=\dfrac{2\sqrt{3}.\sqrt{5}}{\sqrt{2}.\sqrt{5}}=\sqrt{6}\)

Answer 4

\(46,\\ a,VP^2=\left[\sqrt{\dfrac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}}\right]\\ =\dfrac{a+\sqrt{a^2-b}}{2}+\dfrac{a-\sqrt{a^2-b}}{2}\pm2\sqrt{\dfrac{\left(a+\sqrt{a^2-b}\right)\left(a-\sqrt{a^2-b}\right)}{2}}\\ =a\pm\sqrt{b}=\left(\sqrt{a\pm\sqrt{b}}\right)^2\\ \LeftrightarrowĐpcm\)

\(b,VT^2=\left(\sqrt{a+\sqrt{b}}\pm\sqrt{a-\sqrt{b}}\right)^2\\ =a+\sqrt{b}+a-\sqrt{b}\pm2\sqrt{\left(a+\sqrt{b}\right)\left(a-\sqrt{b}\right)}\\ =2a\pm2\sqrt{a^2-b}=2\left(a\pm\sqrt{a-b^2}\right)\\ \LeftrightarrowĐpcm\)

Answer 5

45.

b, \(\left[\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right]:\sqrt{2}\)

\(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{2}\)

\(=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{2}\)

\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{2}=-1\)

Answer 6

45.

d, \(\left[\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\right]:\sqrt{10}\)

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

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

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

\(=1\)