Giải bài 3.3 trang 32

Đề bài

a) ${\left(\sqrt{4,1}\right)}^2-{\left(-\sqrt{6,1}\right)}^2$;

b) ${\left(\sqrt{101}\right)}^2-\sqrt{{\left(-99\right)}^2}$;

c) $\sqrt{{\left(\sqrt{3}+2\sqrt{2}\right)}^2}-\left(-\sqrt{3}+2\sqrt{2}\right)$;

d) $\sqrt{{\left(\sqrt{10}+3\right)}^2}-\sqrt{{\left(\sqrt{10}-3\right)}^2}$.

Lời giải chi tiết

a) ${\left(\sqrt{4,1}\right)}^2-{\left(-\sqrt{6,1}\right)}^2=4,1-6,1=-2$;

b) ${\left(\sqrt{101}\right)}^2-\sqrt{{\left(-99\right)}^2}=101-99=2$;

c) $\sqrt{{\left(\sqrt{3}+2\sqrt{2}\right)}^2}-\left(-\sqrt{3}+2\sqrt{2}\right)$

$=\left|\sqrt{3}+2\sqrt{2}\right|+\sqrt{3}-2\sqrt{2}$

$=\sqrt{3}+2\sqrt{2}+\sqrt{3}-2\sqrt{2}$

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

d) $\sqrt{{\left(\sqrt{10}+3\right)}^2}-\sqrt{{\left(\sqrt{10}-3\right)}^2}$

$=\left|\sqrt{10}+3\right|-\left|\sqrt{10}-3\right|$

$=\sqrt{10}+3-\sqrt{10}+3=6$.