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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 \(x \ge 0,x \ne 1\).
+) Quy đồng mẫu các phân thức.
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\(\begin{array}{l}a)\;\dfrac{{x\sqrt y + y\sqrt x }}{{\sqrt x + \sqrt y }} – \sqrt {xy} \;\;\;\left( {x,\;y > 0} \right)\\ = \dfrac{{\sqrt {xy} \left( {\sqrt x + \sqrt y } \right)}}{{\left( {\sqrt x + \sqrt y } \right)}}\\ = \sqrt {xy} .\end{array}\) \(\begin{array}{l}c)\;\left( {\dfrac{{\sqrt x – 1}}{{\sqrt x + 1}} – \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}}} \right)\left( {\sqrt x – \dfrac{1}{{\sqrt x }}} \right)\;\;\left( {x > 0,\;\;z \ne 1} \right)\\ = \dfrac{{{{\left( {\sqrt x – 1} \right)}^2} – {{\left( {\sqrt x + 1} \right)}^2}}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x – 1} \right)}}.\dfrac{{x – 1}}{{\sqrt x }}\\ = \dfrac{{x – 2\sqrt x + 1 – \left( {x + 2\sqrt x + 1} \right)}}{{x – 1}}.\dfrac{{x – 1}}{{\sqrt x }}\\ = \dfrac{{x – 2\sqrt x + 1 – x – 2\sqrt x – 1}}{{\sqrt x }}\\ = \dfrac{{ – 4\sqrt x }}{{\sqrt x }} = – 4.\end{array}\) |
\(\begin{array}{l}b)\;\left( {2 + \dfrac{{a – \sqrt a }}{{\sqrt a – 1}}} \right)\left( {2 – \dfrac{{a + \sqrt a }}{{\sqrt a + 1}}} \right)\\ = \left( {2 + \dfrac{{\sqrt a \left( {\sqrt a – 1} \right)}}{{\sqrt a – 1}}} \right)\left( {2 – \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a + 1}}} \right)\\ = \left( {2 + \sqrt a } \right)\left( {2 – \sqrt a } \right)\\ = 4 – a.\end{array}\) \(\begin{array}{l}d)\;\;\dfrac{{15\sqrt x – 11}}{{x + 2\sqrt x – 3}} + \dfrac{{3\sqrt x – 2}}{{1 – \sqrt x }} – \dfrac{3}{{\sqrt x + 3}}\;\;\left( {x \ge 0,\;x \ne 1} \right)\\ = \dfrac{{15\sqrt x – 11}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}} – \dfrac{{3\sqrt x – 2}}{{\sqrt x – 1}} – \dfrac{3}{{\sqrt x + 3}}\\ = \dfrac{{15\sqrt x – 11 – \left( {3\sqrt x – 2} \right)\left( {\sqrt x + 3} \right) – 3\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{15\sqrt x – 11 – \left( {3x + 7\sqrt x – 6} \right) – 3\sqrt x + 3}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{12\sqrt x – 8 – 3x – 7\sqrt x + 6}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ = \dfrac{{ – 3x + 5\sqrt x – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ = – \dfrac{{\left( {3\sqrt x – 2} \right)\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ = – \dfrac{{3\sqrt x – 2}}{{\sqrt x + 3}}.\end{array}\) |