Đề bài - bài 6 trang 31 tài liệu dạy – học toán 9 tập 1

Đề bài

Rút gọn :

a) \(\left( {\dfrac{y}{{\sqrt y }} - \dfrac{{\sqrt y }}{{\sqrt y + 1}}} \right):\dfrac{{\sqrt y }}{{y + \sqrt y }}\) với \(y > 0\);

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

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

d) \(\left( {\dfrac{{\sqrt x - \sqrt y }}{{1 + \sqrt {xy} }} + \dfrac{{\sqrt x + \sqrt y }}{{1 - \sqrt {xy} }}} \right):\left( {\dfrac{{x + y + 2xy}}{{1 - xy}} + 1} \right)\)với \(x \ge 0,y \ge 0,xy \ne 1\).

Lời giải chi tiết

\(\begin{array}{l}a)\;\;\left( {\dfrac{y}{{\sqrt y }} - \dfrac{{\sqrt y }}{{\sqrt y + 1}}} \right):\dfrac{{\sqrt y }}{{y + \sqrt y }}\;\;\;\left( {y > 0} \right)\\ = \left( {\sqrt y - \dfrac{{\sqrt y }}{{\sqrt y + 1}}} \right):\dfrac{{\sqrt y }}{{\sqrt y \left( {\sqrt y + 1} \right)}}\\ = \dfrac{{\sqrt y \left( {\sqrt y + 1} \right) - \sqrt y }}{{\sqrt y + 1}}:\dfrac{1}{{\sqrt y + 1}}\\ = \dfrac{{y + \sqrt y - \sqrt y }}{{\sqrt y + 1}}.\left( {\sqrt y + 1} \right)\\ = y.\end{array}\)

\(\begin{array}{l}b)\;\left( {\dfrac{{\sqrt a }}{{\sqrt a - 1}} - \dfrac{1}{{a - \sqrt a }}} \right):\left( {\dfrac{1}{{\sqrt a + 1}} + \dfrac{2}{{a - 1}}} \right)\;\;\;\left( {a > 0,\;\;a \ne 1} \right)\\ = \left( {\dfrac{{\sqrt a }}{{\sqrt a - 1}} - \dfrac{1}{{\sqrt a \left( {\sqrt a - 1} \right)}}} \right):\left( {\dfrac{1}{{\sqrt a + 1}} + \dfrac{2}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}} \right)\\ = \dfrac{{a - 1}}{{\sqrt a \left( {\sqrt a - 1} \right)}}:\dfrac{{\sqrt a - 1 + 2}}{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}\\ = \dfrac{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}}.\dfrac{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}{{\sqrt a + 1}}\\ = \dfrac{{\sqrt a + 1}}{{\sqrt a }}.\dfrac{{\sqrt {a - 1} }}{1} = \dfrac{{a - 1}}{{\sqrt a }}.\end{array}\)

\(\begin{array}{l}c)\;\left( {\dfrac{{x\sqrt x + 1}}{{x\sqrt x + x + \sqrt x + 1}} - \dfrac{{\sqrt x }}{{x + 1}}} \right):\dfrac{{\sqrt x - 1}}{{x + 1}}\;\;\left( {x \ge 0;\;\;x \ne 1} \right)\\ = \left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x\left( {\sqrt x + 1} \right) + \left( {\sqrt x + 1} \right)}} - \dfrac{{\sqrt x }}{{x + 1}}} \right].\dfrac{{x + 1}}{{\sqrt x - 1}}\\ = \left[ {\dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x + 1} \right)}} - \dfrac{{\sqrt x }}{{x + 1}}} \right].\dfrac{{x + 1}}{{\sqrt x - 1}}\\ = \left( {\dfrac{{x - \sqrt x + 1}}{{x + 1}} - \dfrac{{\sqrt x }}{{x + 1}}} \right).\dfrac{{x + 1}}{{\sqrt x - 1}}\\ = \dfrac{{x - \sqrt x + 1 - \sqrt x }}{{x + 1}}.\dfrac{{x + 1}}{{\sqrt x - 1}}\\ = \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x - 1}} = \sqrt x - 1.\end{array}\)

\(\begin{array}{l}d)\;\left( {\dfrac{{\sqrt x - \sqrt y }}{{1 + \sqrt {xy} }} + \dfrac{{\sqrt x + \sqrt y }}{{1 - \sqrt {xy} }}} \right):\left( {\dfrac{{x + y + 2xy}}{{1 - xy}} + 1} \right)\;\;\left( {x \ge 0,\;y \ge 0,\;xy \ne 1} \right)\\ = \dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {1 - \sqrt {xy} } \right) + \left( {\sqrt x + \sqrt y } \right)\left( {1 + \sqrt {xy} } \right)}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}:\dfrac{{x + y + 2xy + 1 - xy}}{{1 - xy}}\\ = \dfrac{{\sqrt x - x\sqrt y + \sqrt y - y\sqrt x + \sqrt x + x\sqrt y + \sqrt y + y\sqrt x }}{{1 - xy}}.\dfrac{{1 - xy}}{{x + y + xy + 1}}\\ = \dfrac{{2\sqrt x + 2\sqrt y }}{{x + y + xy + 1}} = \dfrac{{2\left( {\sqrt x + \sqrt y } \right)}}{{x + 1 + y\left( {x + 1} \right)}}\\ = \dfrac{{2\left( {\sqrt x + \sqrt y } \right)}}{{\left( {x + 1} \right)\left( {y + 1} \right)}}.\end{array}\)