Tháng Tư 19, 2024

Tính : a) \(\frac{5}{{\sqrt 5 – 1}} – \frac{5}{{\sqrt 5 + 1}};\) b) \(\sqrt {{{\left( {\sqrt 5 – 3} \right)}^2}} – \sqrt {\frac{1}{5}} \) A \(\begin{array}{l}{\rm{a)}}\,\,\frac{1}{2}\\{\rm{b)}}\,\,\frac{{15 + 3\sqrt 5 }}{5}\end{array}\) B \(\begin{array}{l}{\rm{a)}}\,\,\frac{3}{2}\\{\rm{b)}}\,\,\frac{{15 – 3\sqrt 5 }}{5}\end{array}\) C \(\begin{array}{l}{\rm{a)}}\,\,\frac{5}{2}\\{\rm{b)}}\,\,\frac{{15 – 6\sqrt 5 }}{5}\end{array}\) D \(\begin{array}{l}{\rm{a)}}\,\,2\\{\rm{b)}}\,\,\frac{{15 + 6\sqrt 5 }}{5}\end{array}\)

Tính :

a) \(\frac{5}{{\sqrt 5 – 1}} – \frac{5}{{\sqrt 5 + 1}};\) b) \(\sqrt {{{\left( {\sqrt 5 – 3} \right)}^2}} – \sqrt {\frac{1}{5}} \)

A \(\begin{array}{l}{\rm{a)}}\,\,\frac{1}{2}\\{\rm{b)}}\,\,\frac{{15 + 3\sqrt 5 }}{5}\end{array}\)

B \(\begin{array}{l}{\rm{a)}}\,\,\frac{3}{2}\\{\rm{b)}}\,\,\frac{{15 – 3\sqrt 5 }}{5}\end{array}\)

C \(\begin{array}{l}{\rm{a)}}\,\,\frac{5}{2}\\{\rm{b)}}\,\,\frac{{15 – 6\sqrt 5 }}{5}\end{array}\)

D \(\begin{array}{l}{\rm{a)}}\,\,2\\{\rm{b)}}\,\,\frac{{15 + 6\sqrt 5 }}{5}\end{array}\)

Hướng dẫn Chọn đáp án là: C

Phương pháp giải:

a) Quy đồng mẫu số và rút gọn biểu thức.

b) Rút gọn căn bậc hai bằng công thức: \(\sqrt {{A^2}} = \left[ \begin{array}{l}A\,\,\,\,khi\,\,\,\,A \ge 0\\ – A\,\,\,khi\,\,\,\,A < 0\end{array} \right.\) và \(\sqrt {\frac{1}{A}} = \frac{{\sqrt A }}{A}\,\,\left( {A > 0} \right)\)

Lời giải chi tiết:

\(\begin{array}{l}a)\,\frac{5}{{\sqrt 5 – 1}} – \frac{5}{{\sqrt 5 + 1}} = \frac{{5\left( {\sqrt 5 + 1} \right) – 5\left( {\sqrt 5 – 1} \right)}}{{\left( {\sqrt 5 + 1} \right)\left( {\sqrt 5 – 1} \right)}}\\ = \frac{{5\sqrt 5 + 5 – 5\sqrt 5 + 5}}{{5 – 1}} = \frac{{10}}{4} = \frac{5}{2}\end{array}\)

\(\begin{array}{l}b)\,\sqrt {{{\left( {\sqrt 5 – 3} \right)}^2}} – \sqrt {\frac{1}{5}} = \left| {\sqrt 5 – 3} \right| – \frac{{\sqrt 5 }}{5}\\ = 3 – \sqrt 5 – \frac{{\sqrt 5 }}{5} = \frac{{15 – 5\sqrt 5 – \sqrt 5 }}{5} = \frac{{15 – 6\sqrt 5 }}{5}\end{array}\)

Chọn C.