Toán Lớp 9: A=(a/a √a-1 + √a/a + √a +1 +1/ 1- √a.
Rút gọn bt A
Tìm giá trị lớn nhất của bt A
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Toán Lớp 9: A=(a/a √a-1 + √a/a + √a +1 +1/ 1- √a.
Rút gọn bt A
Tìm giá trị lớn nhất của bt A
Comments ( 1 )
Dkxd:\left\{ \begin{array}{l}
a \ge 0\\
a\# 1
\end{array} \right.\\
A = \left( {\dfrac{a}{{a\sqrt a – 1}} + \dfrac{{\sqrt a }}{{a + \sqrt a + 1}} + \dfrac{1}{{1 – \sqrt a }}} \right).\dfrac{{\sqrt a – 1}}{2}\\
= \left( {\dfrac{a}{{\left( {\sqrt a – 1} \right)\left( {a + \sqrt a + 1} \right)}} + \dfrac{{\sqrt a }}{{a + \sqrt a + 1}} – \dfrac{1}{{\sqrt a – 1}}} \right)\\
.\dfrac{{\sqrt a – 1}}{2}\\
= \dfrac{{a + \sqrt a \left( {\sqrt a – 1} \right) + a + \sqrt a + 1}}{{\left( {\sqrt a – 1} \right)\left( {a + \sqrt a + 1} \right)}}.\dfrac{{\sqrt a – 1}}{2}\\
= \dfrac{{a + a – \sqrt a + a + \sqrt a + 1}}{{\left( {\sqrt a – 1} \right)\left( {a + \sqrt a + 1} \right)}}.\dfrac{{\sqrt a – 1}}{2}\\
= \dfrac{{3a + 1}}{{2\left( {a + \sqrt a + 1} \right)}}\\
A = \dfrac{{3a + 1}}{{2a + 2\sqrt a + 1}}\\
\Leftrightarrow 2A.a + 2A.\sqrt a + A = 3a + 1\\
\Leftrightarrow \left( {2A – 3} \right).a + 2A.\sqrt a + A – 1 = 0\\
\Delta ‘ \ge 0\\
\Leftrightarrow {A^2} – \left( {2A – 3} \right)\left( {A – 1} \right) \ge 0\\
\Leftrightarrow {A^2} – 2{A^2} + 5A – 3 \ge 0\\
\Leftrightarrow {A^2} – 5A + 3 \le 0\\
\Leftrightarrow \dfrac{{5 – \sqrt {13} }}{2} \le A \le \dfrac{{5 + \sqrt {13} }}{2}\\
\Leftrightarrow GTLN:A = \dfrac{{5 + \sqrt {13} }}{2}
\end{array}$