Question

[Minh tú Trần]

Khử mẫu của căn thức lấy căn

a) \(\sqrt{\frac{y}{5x^3}}\) với x, y cùng dấu; x khác 0

b) \(\sqrt{\frac{5}{x.\left(1-\sqrt{2}\right)}}\)  với x < 0

c) \(\sqrt{\frac{x-1}{2.\left(\sqrt{x}-1\right)}}\)  với x > 1

d) \(a.\sqrt{\frac{4}{a}}\)

e) \(2.\sqrt{\frac{1}{-a}}\)

f) \(\sqrt{\frac{2}{\left(x-1\right)}-\frac{1}{\left(x-1\right)^2}}\)

Answer 1

[Nguyễn Minh Quang]

a. \(\sqrt{\frac{y}{5x^3}}=\sqrt{\frac{5xy}{25x^4}}=\frac{\sqrt{5xy}}{25x^2}\)

b\(\sqrt{\frac{5}{x\left(1-\sqrt{2}\right)}}=\sqrt{\frac{5\times x\left(1+\sqrt{2}\right)}{x^2\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}}=\sqrt{\frac{-5\times x\left(1+\sqrt{2}\right)}{x^2}}=-\frac{\sqrt{-5\times x\left(1+\sqrt{2}\right)}}{x}\)

c.\(\sqrt{\frac{x-1}{2\left(\sqrt{x}-1\right)}}=\sqrt{\frac{\sqrt{x}+1}{2}}=\frac{\sqrt{2\sqrt{x}+2}}{2}\)

d.\(a\sqrt{\frac{4}{a}}=\sqrt{\frac{4a^2}{a}}=\sqrt{4a}=2\sqrt{a}\)

e.\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\left(\text{ do a< 0}\right)\)\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\)( do a <0)

f.\(\sqrt{\frac{2}{x-1}-\frac{1}{\left(x-1\right)^2}}=\sqrt{\frac{2\left(x-1\right)-1}{\left(x-1\right)^2}}=\frac{\sqrt{2x-3}}{\left|x-1\right|}\)