Bài 14 trang 30 Tài liệu dạy – học Toán 9 tập 1

Đề bài

Rút gọn :

a) \(\dfrac{{x\sqrt y  + y\sqrt x }}{{\sqrt x  + \sqrt y }} - \sqrt {xy} \) với \(x > 0,y > 0\); 

b) \(\left( {2 + \dfrac{{a - \sqrt a }}{{\sqrt a  - 1}}} \right)\left( {2 - \dfrac{{a + \sqrt a }}{{\sqrt a  + 1}}} \right)\) với \(a \ge 0,a \ne 1\);

c) \(\left( {\dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}} \right)\left( {\sqrt x  - \dfrac{1}{{\sqrt x }}} \right)\) với \(x > 0,x \ne 1\); 

d) \(\dfrac{{15\sqrt x  - 11}}{{x + 2\sqrt x  - 3}} + \dfrac{{3\sqrt x  - 2}}{{1 - \sqrt x }} - \dfrac{3}{{\sqrt x  + 3}}\) với \(x \ge 0,x \ne 1\).

Lời giải chi tiết

\(\begin{array}{l}a)\;\dfrac{{x\sqrt y  + y\sqrt x }}{{\sqrt x  + \sqrt y }} - \sqrt {xy} \;\;\;\left( {x,\;y > 0} \right)\\ = \dfrac{{\sqrt {xy} \left( {\sqrt x  + \sqrt y } \right)}}{{\left( {\sqrt x  + \sqrt y } \right)}}\\ = \sqrt {xy} .\end{array}\)

\(\begin{array}{l}b)\;\left( {2 + \dfrac{{a - \sqrt a }}{{\sqrt a  - 1}}} \right)\left( {2 - \dfrac{{a + \sqrt a }}{{\sqrt a  + 1}}} \right)\\ = \left( {2 + \dfrac{{\sqrt a \left( {\sqrt a  - 1} \right)}}{{\sqrt a  - 1}}} \right)\left( {2 - \dfrac{{\sqrt a \left( {\sqrt a  + 1} \right)}}{{\sqrt a  + 1}}} \right)\\ = \left( {2 + \sqrt a } \right)\left( {2 - \sqrt a } \right)\\ = 4 - a.\end{array}\)

\(\begin{array}{l}c)\;\left( {\dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} - \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}} \right)\left( {\sqrt x  - \dfrac{1}{{\sqrt x }}} \right)\;\;\left( {x > 0,\;\;z \ne 1} \right)\\ = \dfrac{{{{\left( {\sqrt x  - 1} \right)}^2} - {{\left( {\sqrt x  + 1} \right)}^2}}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  - 1} \right)}}.\dfrac{{x - 1}}{{\sqrt x }}\\ = \dfrac{{x - 2\sqrt x  + 1 - \left( {x + 2\sqrt x  + 1} \right)}}{{x - 1}}.\dfrac{{x - 1}}{{\sqrt x }}\\ = \dfrac{{x - 2\sqrt x  + 1 - x - 2\sqrt x  - 1}}{{\sqrt x }}\\ = \dfrac{{ - 4\sqrt x }}{{\sqrt x }} =  - 4.\end{array}\)

\(\begin{array}{l}d)\;\;\dfrac{{15\sqrt x  - 11}}{{x + 2\sqrt x  - 3}} + \dfrac{{3\sqrt x  - 2}}{{1 - \sqrt x }} - \dfrac{3}{{\sqrt x  + 3}}\;\;\left( {x \ge 0,\;x \ne 1} \right)\\ = \dfrac{{15\sqrt x  - 11}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}} - \dfrac{{3\sqrt x  - 2}}{{\sqrt x  - 1}} - \dfrac{3}{{\sqrt x  + 3}}\\ = \dfrac{{15\sqrt x  - 11 - \left( {3\sqrt x  - 2} \right)\left( {\sqrt x  + 3} \right) - 3\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{15\sqrt x  - 11 - \left( {3x + 7\sqrt x  - 6} \right) - 3\sqrt x  + 3}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{12\sqrt x  - 8 - 3x - 7\sqrt x  + 6}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\ = \dfrac{{ - 3x + 5\sqrt x  - 2}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\ =  - \dfrac{{\left( {3\sqrt x  - 2} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 3} \right)}}\\ =  - \dfrac{{3\sqrt x  - 2}}{{\sqrt x  + 3}}.\end{array}\)