a: Xét (O) có

MA là tiếp tuyến

MB là tiếp tuyến

Do đó: MA=MB

hay M nằm trên đường trung trực của AB(1)

Ta có: OA=OB

nên O nằm trên đường trung trực của AB(2)

Từ (1) và (2) suy ra OM⊥AB

Bài 4:

a: \(\sqrt{1,6}\cdot\sqrt{250}+\sqrt{19,6}:\sqrt{4,9}\)

\(=\sqrt{1,6\cdot250}+\sqrt4\)

\(=\sqrt{400}+2=20+2=22\)

b: \(\sqrt{1\frac34\cdot2\frac27\cdot5\frac49}\)

\(=\sqrt{\frac74\cdot\frac{16}{7}\cdot\frac{49}{9}}=\sqrt{\frac{16}{4}\cdot\frac{49}{9}}=2\cdot\frac73=\frac{14}{3}\)

c: \(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)

\(=20\sqrt{20}-15\sqrt{45}+5\sqrt5\)

\(=40\sqrt5-45\sqrt5+5\sqrt5=0\)

d: \(\left(\sqrt{325}-\sqrt{117}+2\sqrt{208}\right):\sqrt{13}\)

\(=\sqrt{25}-\sqrt9+2\cdot\sqrt{16}\)

\(=5-3+2\cdot4\)

=2+8

=10

e: \(\frac{2\sqrt8-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt5+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\frac{4\sqrt2-2\sqrt3}{\sqrt6\left(\sqrt3-2\sqrt2\right)}-\frac{\sqrt5+\sqrt{27}}{\sqrt6\left(\sqrt5+\sqrt{27}\right)}\)

\(=\frac{2\left(2\sqrt2-\sqrt3\right)}{-\sqrt6\left(2\sqrt2-\sqrt3\right)}-\frac{1}{\sqrt6}=-\frac{2}{\sqrt6}-\frac{1}{\sqrt6}=-\frac{3}{\sqrt6}=\frac{-3\sqrt6}{6}=-\frac{\sqrt6}{2}\)

f: \(\frac{3+2\sqrt3}{\sqrt3}+\frac{2+\sqrt2}{\sqrt2+1}-\left(\sqrt2+\sqrt3\right)\)

\(=2+\sqrt3+\frac{\sqrt2\left(\sqrt2+1\right)}{\sqrt2+1}-\sqrt2-\sqrt3\)

\(=2-\sqrt2+\sqrt2\)

=2

bài 1:

\(a.\sqrt{25 . 144}=\sqrt{25}.\sqrt{144}=5.12=60\)

\(b.\sqrt{45 . 80}=\sqrt{9 . 5 . 5 . 16}=\sqrt{9 . 25 . 16}=\sqrt{9}.\sqrt{25}.\sqrt{16}=3.5.4=60\)

\(c.\sqrt{52}.\sqrt{13}=\sqrt{52 . 13}=\sqrt{4 . 13 . 13}=\sqrt{4 . 13^2}=\sqrt{4}.\sqrt{13^2}=2.13=26\)

\(d.\sqrt{7}.\sqrt{28}=\sqrt{7 . 28}=\sqrt{7 . 7 . 4}=\sqrt{7^2 . 4}=\sqrt{7^2}.\sqrt{4}=7.2=14\)

\(e.\sqrt{1 \frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{\sqrt{25}}{\sqrt{16}}=\frac{5}{4}\)

\(f.\sqrt{\frac{25}{64}}=\frac{\sqrt{25}}{\sqrt{64}}=\frac{5}{8}\)

\(g.\frac{\sqrt{12,5}}{\sqrt{0,5}}=\sqrt{\frac{12,5}{0,5}}=\sqrt{25}=5\)

\(h.\frac{\sqrt{230}}{\sqrt{2,3}}=\sqrt{\frac{230}{2,3}}=\sqrt{100}=10\)

bài 2:

\(a.\left(\sqrt{\frac{2}{3}}+\sqrt{\frac{50}{3}}-\sqrt{24}\right).\sqrt{6}\)

\(= \sqrt{\frac{2}{3}} . \sqrt{6} + \sqrt{\frac{50}{3}} . \sqrt{6} - \sqrt{24} . \sqrt{6}\)

\(= \sqrt{\frac{2}{3} . 6} + \sqrt{\frac{50}{3} . 6} - \sqrt{24 . 6}\)

\(= \sqrt{4} + \sqrt{100} - \sqrt{144}\)

\(=2+10-12=0\)

b. \(\sqrt{3 + \sqrt{5}}.\sqrt{2}=\sqrt{(3 + \sqrt{5}) . 2}\)

\(=\sqrt{6 + 2\sqrt{5}}=\sqrt{5 + 2\sqrt{5} + 1}\)

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

\(c.\left(\sqrt{\frac{3}{4}}-\sqrt{3}+5\sqrt{\frac{4}{3}}\right).\sqrt{12}\)

\(= \sqrt{\frac{3}{4}} . \sqrt{12} - \sqrt{3} . \sqrt{12} + 5\sqrt{\frac{4}{3}} . \sqrt{12}\)

\(= \sqrt{\frac{3}{4} . 12} - \sqrt{3 . 12} + 5\sqrt{\frac{4}{3} . 12}\)

\(= \sqrt{9} - \sqrt{36} + 5\sqrt{16}\)

\(= 3 - 6 + 5 . 4\)

\(=3-6+20=17\)

\(d.\sqrt{3 - \sqrt{5}}.\sqrt{8}=\sqrt{3 - \sqrt{5}}.\sqrt{2}.\sqrt{4}\)

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

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

\(=2\vert{}\sqrt{5}-1\vert{}=2(\sqrt{5}-1)=2\sqrt{5}-2\)

bài 3:

\(a.\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)

\(= \sqrt{\frac{1}{7}} : \sqrt{7} - \sqrt{\frac{16}{7}} : \sqrt{7} + \sqrt{7} : \sqrt{7}\)

\(= \sqrt{\frac{1}{7} : 7} - \sqrt{\frac{16}{7} : 7} + 1\)

\(= \sqrt{\frac{1}{49}} - \sqrt{\frac{16}{49}} + 1\)

\(=\frac{1}{7}-\frac{4}{7}+1=-\frac{3}{7}+1=\frac{4}{7}\)

\(b.\sqrt{36 - 12\sqrt{5}}:\sqrt{6}=\sqrt{\frac{36 - 12\sqrt{5}}{6}}\)

\(=\sqrt{6 - 2\sqrt{5}}=\sqrt{5 - 2\sqrt{5} + 1}\)

\(=\sqrt{(\sqrt{5} - 1)^2}=\vert{}\sqrt{5}-1\vert{}=\sqrt{5}-1\)

\(c.\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)

\(= \sqrt{\frac{1}{3}} : \sqrt{3} - \sqrt{\frac{4}{3}} : \sqrt{3} + \sqrt{3} : \sqrt{3}\)

\(= \sqrt{\frac{1}{3} : 3} - \sqrt{\frac{4}{3} : 3} + 1\)

\(=\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+1=\frac{1}{3}-\frac{2}{3}+1\)

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

\(e.\sqrt{3 - \sqrt{5}}:\sqrt{2}=\sqrt{\frac{3 - \sqrt{5}}{2}}\)

\(=\sqrt{\frac{6 - 2\sqrt{5}}{4}}=\frac{\sqrt{6 - 2\sqrt{5}}}{\sqrt{4}}\)

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

\(=\frac{\vert{}\sqrt{5} - 1\vert{}}{2}=\frac{\sqrt{5} - 1}{2}\)

bài 4:

\(a.\sqrt{1,6}.\sqrt{250}+\sqrt{19,6}:\sqrt{4,9}=\sqrt{1,6 . 250}+\sqrt{\frac{19,6}{4,9}}\)

\(=\sqrt{400}+\sqrt{4}=20+2=22\)

\(b.\sqrt{1 \frac{3}{4}}.\sqrt{2 \frac{2}{7}}.\sqrt{5 \frac{4}{9}}=\sqrt{\frac{7}{4}}.\sqrt{\frac{16}{7}}.\sqrt{\frac{49}{9}}\)

\(=\sqrt{\frac{7}{4} . \frac{16}{7} . \frac{49}{9}}=\sqrt{\frac{16 . 49}{4 . 9}}=\sqrt{\frac{4 . 49}{9}}\)

\(=\frac{\sqrt{4} . \sqrt{49}}{\sqrt{9}}=\frac{2 . 7}{3}=\frac{14}{3}\)

\(c.\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)

\(= 20\sqrt{300} : \sqrt{15} - 15\sqrt{675} : \sqrt{15} + 5\sqrt{75} : \sqrt{15}\)

\(= 20\sqrt{\frac{300}{15}} - 15\sqrt{\frac{675}{15}} + 5\sqrt{\frac{75}{15}}\)

\(= 20\sqrt{20} - 15\sqrt{45} + 5\sqrt{5}\)

\(= 20\sqrt{4 . 5} - 15\sqrt{9 . 5} + 5\sqrt{5}\)

\(= 20 . 2\sqrt{5} - 15 . 3\sqrt{5} + 5\sqrt{5}\)

\(= 40\sqrt{5} - 45\sqrt{5} + 5\sqrt{5}\)

\(= (40 - 45 + 5)\sqrt{5}\)

\(=0\sqrt{5}=0\)

d. \(\left( \sqrt{325} - \sqrt{117} + 2\sqrt{208} \right) : \sqrt{13}\)

\(= \sqrt{325} : \sqrt{13} - \sqrt{117} : \sqrt{13} + 2\sqrt{208} : \sqrt{13}\)

\(= \sqrt{\frac{325}{13}} - \sqrt{\frac{117}{13}} + 2\sqrt{\frac{208}{13}}\)

\(= \sqrt{25} - \sqrt{9} + 2\sqrt{16}\)

\(=5-3+2.4=10\)

\(e.\frac{2\sqrt{8} - \sqrt{12}}{\sqrt{18} - \sqrt{48}}.\frac{\sqrt{5} + \sqrt{27}}{\sqrt{30} + \sqrt{162}}=\frac{2\sqrt{4 . 2} - \sqrt{4 . 3}}{\sqrt{9 . 2} - \sqrt{16 . 3}}.\frac{\sqrt{5} + \sqrt{27}}{\sqrt{6 . 5} + \sqrt{81 . 2}}\)

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

\(=\frac{2(2\sqrt{2} - \sqrt{3})}{3\sqrt{2} - 4\sqrt{3}}.\frac{1}{\sqrt{6}}=\frac{2(2\sqrt{2} - \sqrt{3})}{\sqrt{6}(3\sqrt{2} - 4\sqrt{3})}\)