Thì gọi em là cẩu sư đệ

Báo cáo=))
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1, \(\sqrt{8+2\sqrt{15}}=\sqrt{8+2\sqrt{5.3}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)

2, \(\sqrt{15-2\sqrt{14}}=\sqrt{14-2\sqrt{14}+1}=\sqrt{\left(\sqrt{14}-1\right)^2}=\sqrt{14}-1\)

3, \(\sqrt{21+8\sqrt{5}}=\sqrt{21+2.4\sqrt{5}}=\sqrt{16+2.4\sqrt{5}+5}\)

\(=\sqrt{\left(4+\sqrt{5}\right)^2}=4+\sqrt{5}\)

Xét tg MAC và tg MCB có

\(\widehat{BMC}\) chung

\(sd\widehat{MCA}=\frac{1}{2}sd\)cung AC (góc giữa tiếp tuyến và dây cung)

\(sd\widehat{MBC}=\frac{1}{2}sd\) cung AC (góc nội tiếp đường tròn)

\(\Rightarrow\widehat{MCA}=\widehat{MBC}\)

=> tg MAC đồng dạng tg MCB (g.g.g) \(\Rightarrow\frac{MC}{MB}=\frac{MA}{MC}\Rightarrow MC^2=MA.MB\)

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a, Ta có: ∆AEF ~ ∆MCE (c.g.c)

=>  A F E ^ = A C B ^

b, Ta có: ∆MFB ~ ∆MCE (g.g)

=> ME.MF = MB.MC

a, Ta có: ∆AEF ~ ∆MCE (c.g.c)

=>  A F E ^ = A C B ^

b, Ta có: ∆MFB ~ ∆MCE (g.g)

=> ME.MF = MB.MC

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

b, \(\sqrt{\left(\sqrt{7}-3\right)^2}-\sqrt{\left(4-\sqrt{7}\right)^2}=\sqrt{7}-3-4+\sqrt{7}=2\sqrt{7}-7\)

c, \(\sqrt{9-4\sqrt{5}}+\sqrt{14-6\sqrt{5}}\sqrt{5-2.2\sqrt{5}+4}+\sqrt{9-2.3\sqrt{5}+5}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}=\sqrt{5}-2+3-\sqrt{5}=1\)

d;e;f tương tự