Rút gọn B. Sau đó so sánh B và 1:
B = ( √x + √y / 2√x – 2√y – 2√xy / x-y) . 2√x / √x – √y (với x,y >0)
0 bình luận về “Rút gọn B. Sau đó so sánh B và 1:
B = ( √x + √y / 2√x – 2√y – 2√xy / x-y) . 2√x / √x – √y (với x,y >0)”
Giải thích các bước giải:
Ta có:
\(\begin{array}{l} B = \left( {\dfrac{{\sqrt x + \sqrt y }}{{2\sqrt x – 2\sqrt y }} – \dfrac{{2\sqrt {xy} }}{{x – y}}} \right).\dfrac{{2\sqrt x }}{{\sqrt x – \sqrt y }}\\ = \left( {\dfrac{{\sqrt x + \sqrt y }}{{2.\left( {\sqrt x – \sqrt y } \right)}} – \dfrac{{2\sqrt {xy} }}{{\left( {\sqrt x – \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right).\dfrac{{2\sqrt x }}{{\sqrt x – \sqrt y }}\\ = \dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2} – 4.\sqrt {xy} }}{{2\left( {\sqrt x – \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}.\dfrac{{2\sqrt x }}{{\sqrt x – \sqrt y }}\\ = \dfrac{{{{\left( {\sqrt x – \sqrt y } \right)}^2}}}{{2.\left( {\sqrt x – \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}.\dfrac{{2\sqrt x }}{{\sqrt x – \sqrt y }}\\ = \dfrac{{\sqrt x }}{{\sqrt x + \sqrt y }}\\ B – 1 = \dfrac{{\sqrt x }}{{\sqrt x + \sqrt y }} – 1 = \dfrac{{\sqrt x – \left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x + \sqrt y }} = \dfrac{{ – \sqrt y }}{{\sqrt x + \sqrt y }} < 0,\,\,\,\forall x,y > 0\\ \Rightarrow B < 1 \end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
B = \left( {\dfrac{{\sqrt x + \sqrt y }}{{2\sqrt x – 2\sqrt y }} – \dfrac{{2\sqrt {xy} }}{{x – y}}} \right).\dfrac{{2\sqrt x }}{{\sqrt x – \sqrt y }}\\
= \left( {\dfrac{{\sqrt x + \sqrt y }}{{2.\left( {\sqrt x – \sqrt y } \right)}} – \dfrac{{2\sqrt {xy} }}{{\left( {\sqrt x – \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}} \right).\dfrac{{2\sqrt x }}{{\sqrt x – \sqrt y }}\\
= \dfrac{{{{\left( {\sqrt x + \sqrt y } \right)}^2} – 4.\sqrt {xy} }}{{2\left( {\sqrt x – \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}.\dfrac{{2\sqrt x }}{{\sqrt x – \sqrt y }}\\
= \dfrac{{{{\left( {\sqrt x – \sqrt y } \right)}^2}}}{{2.\left( {\sqrt x – \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}.\dfrac{{2\sqrt x }}{{\sqrt x – \sqrt y }}\\
= \dfrac{{\sqrt x }}{{\sqrt x + \sqrt y }}\\
B – 1 = \dfrac{{\sqrt x }}{{\sqrt x + \sqrt y }} – 1 = \dfrac{{\sqrt x – \left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x + \sqrt y }} = \dfrac{{ – \sqrt y }}{{\sqrt x + \sqrt y }} < 0,\,\,\,\forall x,y > 0\\
\Rightarrow B < 1
\end{array}\)