Đề bài - bài 6 trang 29 tài liệu dạy – học toán 9 tập 1

Đề bài

Trục căn thức ở mẫu ( giả thiết các biểu thức chữ đều có nghĩa ) :

a) \(\dfrac{7}{{\sqrt 3 }};\;\dfrac{3}{{2\sqrt 5 }};\;\dfrac{5}{{3\sqrt {12} }};\;\dfrac{2}{{3\sqrt {20} }}\);

b) \(\dfrac{{\sqrt 3 + 3}}{{5\sqrt 3 }};\;\dfrac{{7 - \sqrt 7 }}{{\sqrt 7 - 1}}\);

c) \(\dfrac{5}{{\sqrt 5 + 2}};\;\dfrac{3}{{\;\sqrt 3 - 1}};\;\dfrac{2}{{\sqrt 5 + \sqrt 3 }};\)\(\;\dfrac{{\sqrt 5 + 2}}{{\sqrt 5 - 2}};\;\dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }}\);

d) \(\dfrac{{y + a\sqrt y }}{{a\sqrt y }};\;\dfrac{{b - \sqrt b }}{{\sqrt b - 1}}\);

e) \(\dfrac{b}{{5 + \sqrt b }};\;\dfrac{p}{{2\sqrt p - 1}};\;\dfrac{{\sqrt a + \sqrt b }}{{\sqrt a - \sqrt b }}\).

Lời giải chi tiết

\(\begin{array}{l}a)\;\;\dfrac{7}{{\sqrt 3 }} = \dfrac{{7\sqrt 3 }}{3};\\\dfrac{3}{{2\sqrt 5 }} = \dfrac{{3\sqrt 5 }}{{2.5}} = \dfrac{{3\sqrt 5 }}{{10}}\\\;\;\;\dfrac{5}{{3\sqrt {12} }} = \dfrac{5}{{3\sqrt {{2^2}.3} }} = \dfrac{5}{{6\sqrt 3 }} = \dfrac{{5\sqrt 3 }}{{18}}\\\;\;\;\dfrac{2}{{3\sqrt {20} }} = \dfrac{2}{{3\sqrt {{2^2}.5} }} = \dfrac{2}{{6\sqrt 5 }} = \dfrac{{2\sqrt 5 }}{{30}} = \dfrac{{\sqrt 5 }}{{15}}.\end{array}\)

\(\begin{array}{l}b)\;\dfrac{{\sqrt 3 + 3}}{{5\sqrt 3 }} = \dfrac{{\sqrt 3 \left( {1 + \sqrt 3 } \right)}}{{5\sqrt 3 }} = \dfrac{{\sqrt 3 + 1}}{5}\\\;\;\;\dfrac{{7 - \sqrt 7 }}{{\sqrt 7 - 1}} = \dfrac{{\sqrt 7 \left( {\sqrt 7 - 1} \right)}}{{\sqrt 7 - 1}} = \sqrt 7 .\end{array}\)

\(\begin{array}{l}c)\;\;\dfrac{5}{{\sqrt 5 + 2}} = \dfrac{{5\left( {\sqrt 5 - 2} \right)}}{{5 - {2^2}}} \\\;\;\;= \dfrac{{5\sqrt 5 - 10}}{{5 - 4}} = 5\sqrt 5 - 10\\\;\;\;\;\;\dfrac{3}{{\sqrt 3 - 1}} = \dfrac{{3\left( {\sqrt 3 + 1} \right)}}{{3 - 1}} = \dfrac{{3\sqrt 3 + 3}}{2}\\\;\;\;\;\;\dfrac{2}{{\sqrt 5 + \sqrt 3 }} = \dfrac{{2\left( {\sqrt 5 - \sqrt 3 } \right)}}{{5 - 3}} \\\;\;\;= \dfrac{{2\left( {\sqrt 5 - \sqrt 3 } \right)}}{2} = \sqrt 5 - \sqrt 3 \\\;\;\;\;\dfrac{{\sqrt 5 + 2}}{{\sqrt 5 - 2}} = \dfrac{{{{\left( {\sqrt 5 + 2} \right)}^2}}}{{5 - {2^2}}} \\\;\;\;= \dfrac{{5 + 2.2\sqrt 5 + {2^2}}}{1} = 9 + 4\sqrt 5 \\\;\;\;\;\dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }} = \dfrac{{{{\left( {\sqrt 7 - \sqrt 5 } \right)}^2}}}{{7 - 5}}\\\;\;\; = \dfrac{{7 - 2.\sqrt {5.7} + 5}}{2} = 6 - \sqrt {35} .\end{array}\)

\(\begin{array}{l}d)\;\dfrac{{y + a\sqrt y }}{{a\sqrt y }} = \dfrac{{\sqrt y \left( {\sqrt y + a} \right)}}{{a\sqrt y }} = \dfrac{{\sqrt y + a}}{a}\\\;\;\;\dfrac{{b - \sqrt b }}{{\sqrt b - 1}} = \dfrac{{\sqrt b \left( {\sqrt b - 1} \right)}}{{\sqrt b - 1}} = \sqrt b .\end{array}\)

\(\begin{array}{l}e)\;\dfrac{b}{{5 + \sqrt b }} = \dfrac{{b\left( {\sqrt b - 5} \right)}}{{b - {5^2}}} = \dfrac{{b\left( {\sqrt b - b} \right)}}{{b - 25}}\\\;\;\;\;\dfrac{p}{{2\sqrt p - 1}} = \dfrac{{p\left( {2\sqrt p + 1} \right)}}{{{{\left( {2\sqrt p } \right)}^2} - 1}} = \dfrac{{2p\sqrt p + p}}{{4p - 1}}\\\;\;\;\;\dfrac{{a + \sqrt b }}{{\sqrt a - \sqrt b }} = \dfrac{{\left( {a + \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}\end{array}\)