Toán Lớp 9: Bài 5: Cho Q = (1/√????−1−1/√????):(√????+1/√????−2−√????+2/√????−1) ( ????>0;????≠4;????≠1)
a)Rút gọn
b)Tính Q khi a = 9
c)Tìm a để Q dương.
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Toán Lớp 9: Bài 5: Cho Q = (1/√????−1−1/√????):(√????+1/√????−2−√????+2/√????−1) ( ????>0;????≠4;????≠1)
a)Rút gọn
b)Tính Q khi a = 9
c)Tìm a để Q dương.
Comments ( 1 )
a)Dkxd:a > 0;a \ne 1;a \ne 4\\
Q = \left( {\dfrac{1}{{\sqrt a – 1}} – \dfrac{1}{{\sqrt a }}} \right):\left( {\dfrac{{\sqrt a + 1}}{{\sqrt a – 2}} – \dfrac{{\sqrt a + 2}}{{\sqrt a – 1}}} \right)\\
= \dfrac{{\sqrt a – \left( {\sqrt a – 1} \right)}}{{\sqrt a \left( {\sqrt a – 1} \right)}}\\
:\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a – 1} \right) – \left( {\sqrt a + 2} \right)\left( {\sqrt a – 2} \right)}}{{\left( {\sqrt a – 1} \right)\left( {\sqrt a – 2} \right)}}\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a – 1} \right)}}.\dfrac{{\left( {\sqrt a – 1} \right)\left( {\sqrt a – 2} \right)}}{{a – 1 – a + 4}}\\
= \dfrac{1}{{\sqrt a }}.\dfrac{{\sqrt a – 2}}{3}\\
= \dfrac{{\sqrt a – 2}}{{3\sqrt a }}\\
b)a = 9\left( {tmdk} \right)\\
\Leftrightarrow \sqrt a = 3\\
\Leftrightarrow Q = \dfrac{{3 – 2}}{{3.3}} = \dfrac{1}{9}\\
c)Q > 0\\
\Leftrightarrow \dfrac{{\sqrt a – 2}}{{3\sqrt a }} > 0\\
\Leftrightarrow \sqrt a – 2 > 0\\
\Leftrightarrow \sqrt a > 2\\
\Leftrightarrow a > 4\\
Vậy\,a > 4
\end{array}$