Question

đưa nhân tử ra ngoài dấu căn:

a, \(\sqrt{5\left(1-\sqrt{2}\right)^2}\)

b, \(\sqrt{27\left(2-\sqrt{5}\right)^2}\)

c, \(\sqrt{\dfrac{2}{\left(3-\sqrt{10}\right)^2}}\)

d, \(\sqrt{\dfrac{5\left(1-\sqrt{3}\right)^2}{4}}\)

Answer 1

a, \(\sqrt{5\left(1-\sqrt{2}\right)^2}=\sqrt{5}.\sqrt{\left(1-\sqrt{2}\right)^2}\)

\(=\sqrt{5}.\left(1-\sqrt{2}\right)=\sqrt{5}-\sqrt{5}.\sqrt{2}=\sqrt{5}-\sqrt{10}\)

b, \(\sqrt{27\left(2-\sqrt{5}\right)^2}=\sqrt{27}.\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(=\sqrt{27}.\left(2-\sqrt{5}\right)=2\sqrt{27}-\sqrt{135}\)

c, \(\sqrt{\dfrac{2}{\left(3-\sqrt{10}\right)^2}}=\dfrac{\sqrt{2}}{\sqrt{\left(3-\sqrt{10}\right)^2}}\)

\(=\dfrac{\sqrt{2}}{3-\sqrt{10}}\)

d, \(\sqrt{\dfrac{5\left(1-\sqrt{3}\right)^2}{4}}=\dfrac{\sqrt{5\left(1-\sqrt{3}\right)^2}}{\sqrt{4}}\)

\(=\dfrac{\sqrt{5}.\left(1-\sqrt{3}\right)}{2}=\dfrac{\sqrt{5}-\sqrt{15}}{2}\)

Chúc bạn học tốt!!!

Answer 2

a) \(\sqrt{5\left(1-\sqrt{2}\right)^2}\)

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

= \(\sqrt{5}.\left(\sqrt{2}-1\right)\)

= \(\sqrt{10}-\sqrt{5}\)

b) \(\sqrt{27\left(2-\sqrt{5}\right)^2}\)

= \(\sqrt{27}.\sqrt{\left(2-\sqrt{5}\right)^2}\)

= \(\sqrt{27}.\left(\sqrt{5}-2\right)\)

= \(\sqrt{135}-2\sqrt{27}\)

c) \(\sqrt{\dfrac{2}{\left(3-\sqrt{10}\right)^2}}\)

= \(\dfrac{\sqrt{2}}{\sqrt{\left(3-\sqrt{10}\right)^2}}\)

= \(\dfrac{\sqrt{2}}{\sqrt{10}-3}\)

d) \(\sqrt{\dfrac{5\left(1-\sqrt{3}\right)^2}{4}}\)

= \(\dfrac{\sqrt{5}.\sqrt{\left(1-\sqrt{3}\right)^2}}{\sqrt{4}}\)

= \(\dfrac{\sqrt{5}.\left(\sqrt{3}-1\right)}{2}\)

= \(\dfrac{\sqrt{15}-\sqrt{5}}{2}\)