Question

Bai 1: Rut gon:

a) \(\sqrt{25^2-24^2}\)

b) \(\sqrt{21.8^2-18.2^2}\)

bai 2: Tinh:

a) \(\sqrt{\sqrt{3}+\sqrt{2}}.\sqrt{\sqrt{3}-\sqrt{2}}\)

Answer 1

\(a\text{)}\:\sqrt{25^2-24^2}=\text{ }\sqrt{\left(25-24\right)\left(25+24\right)}=\sqrt{49}=7\)

\(b\text{)}\:\sqrt{21.8^2-18.2^2}=\text{ }\sqrt{3.7.2^6-2^3.3^2}\\ =\sqrt{3.2^3\left(7.2^3-3\right)}=\sqrt{24.53}=\sqrt{1272}=2\sqrt{318}\)

Answer 2

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

Answer 3

Bài 1 :

a , \(\sqrt{25^2-24^2}=\sqrt{\left(25-24\right)\left(25+24\right)}=\sqrt{49}=7\)

b , \(\sqrt{21.8^2-18.2^2}=\) \(\sqrt{21.4^2.2^3-18.2^2}=\sqrt{2^2\left(21.4^2-18\right)}=2\sqrt{6\left(7.8-3\right)}=2\sqrt{6.53}=2\sqrt{318}\)

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