Giải bài tập 3.10 trang 58

Đề bài

Tính nhanh:
a) $\sqrt{{37}^2-{12}^2}$

b) $\sqrt{{101}^2-{20}^2}$

c) $\sqrt{\sqrt{7}-\sqrt{3}}.\sqrt{\sqrt{7}+\sqrt{3}}$

Lời giải chi tiết

a) $\sqrt{{37}^2-{12}^2}$$=\sqrt{\left(37-12\right)\left(37+12\right)}$$=\sqrt{25.49}$$=\sqrt{25}.\sqrt{49}$$=5.7$$=35$.

b) $\sqrt{{101}^2-{20}^2}$$=\sqrt{\left(101-20\right)\left(101+20\right)}$$=\sqrt{81.121}$$=\sqrt{81}.\sqrt{121}$$=9.11$$=99$.

c) $\sqrt{\sqrt{7}-\sqrt{3}}.\sqrt{\sqrt{7}+\sqrt{3}}$$=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}$$=\sqrt{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{3}\right)}^2}$$=\sqrt{7-3}$$=\sqrt{4}$$=2$.