Question

1, tính

a, \(\sqrt{10^2-6^2}\) b, \(\sqrt{17^2-8^2}\) c, \(\sqrt{2,9^2-2,1^2}\)

Answer 1

a)\(\sqrt{10^2-6^2}\)=\(\sqrt{\left(10-6\right)\left(10+6\right)}\)

=\(\sqrt{4\times16}\)=\(2\times4\) = 8

b)\(\sqrt{17^2-8^2}\)=\(\sqrt{\left(17-8\right)\left(17+8\right)}\)=\(\sqrt{9\times25}\)

=\(3\times5=15\)

c)\(\sqrt{2,9^2-2.1^2}\)=\(\sqrt{\left(2,9-2,1\right)\left(2,9+2,1\right)}\)=\(\sqrt{0,8\times5}\)=\(\sqrt{4}\)= 2

Answer 2

a) \(\sqrt{10^2-6^2}=\sqrt{\left(10-6\right)\left(10+6\right)}=\sqrt{4.16}=\sqrt{4}.\sqrt{16}=2.4=8\)b) \(\sqrt{17^2-8^2}=\sqrt{\left(17-8\right)\left(17+8\right)}=\sqrt{9.25}=\sqrt{9}.\sqrt{25}=3.5=15\)

c) \(\sqrt{2,9^2-2,1^2}=\sqrt{\left(2,9-2,1\right)\left(2,9+2,1\right)}=\sqrt{0,8.5}=\sqrt{\dfrac{8}{10}.5}=\sqrt{\dfrac{40}{10}}=\sqrt{4}=2\)