Question

[Bảo Vi]

Rút gọn các biểu thức sau:

\(a.\frac{1}{\sqrt{2}-\sqrt{3}}-\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)     \(b.\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)

\(c.\sqrt{\frac{3+2\sqrt{2}}{3-2\sqrt{2}}}+\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\)                \(d.\frac{1}{\sqrt{5}-2}.\sqrt{\frac{2\sqrt{5}-4}{2\sqrt{5}+4}}\)

\(e.x+1-\sqrt{x^2-2x+1}\left(x>=1\right)\)       \(f.3x+\sqrt{9x^2+6x+1}\left(x< \frac{1}{3}\right)\)

\(g.\frac{1}{9x^2-1}.\sqrt{1-6x+9x^2}\left(x< =\frac{1}{3}\right)\)        \(h.\frac{a-b}{3b}.\sqrt{\frac{4a^2b^4}{a^2-2ab+b^2}}\left(a< b< 0\right)\)

Answer 1

[vo phi hung]

b)

)\(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)

= \(\frac{2}{2-\sqrt{5}}-\frac{2}{2+\sqrt{5}}\)

=\(\frac{2\left(2+\sqrt{5}\right)-2\left(2-\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}\)

=\(\frac{4+2\sqrt{5}-4+2\sqrt{5}}{2^2-\sqrt{5}^2}\)

=\(\frac{4\sqrt{5}}{4-5}\)

=\(\frac{4\sqrt{5}}{-1}\)

\(-4\sqrt{5}\)