Toán Lớp 9: Cho A= (căn a/căn a-1 – 1/a-căn a) : (1/căn a+1 + 2/a-1)
a) Rút gọn A
b) Tính giá trị A khi a = 3+2 căn 2
c) Tìm giá trị của a sao cho K < 0
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Toán Lớp 9: Cho A= (căn a/căn a-1 – 1/a-căn a) : (1/căn a+1 + 2/a-1)
a) Rút gọn A
b) Tính giá trị A khi a = 3+2 căn 2
c) Tìm giá trị của a sao cho K < 0
Comments ( 1 )
a)Dkxd:a > 0;a\# 1\\
A = \left( {\dfrac{{\sqrt a }}{{\sqrt a – 1}} – \dfrac{1}{{a – \sqrt a }}} \right):\left( {\dfrac{1}{{\sqrt a + 1}} + \dfrac{2}{{a – 1}}} \right)\\
= \dfrac{{\sqrt a .\sqrt 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{{a – 1}}{{\sqrt a \left( {\sqrt a – 1} \right)}}.\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a – 1} \right)}}{{a + 1}}\\
= \dfrac{{a – 1}}{{\sqrt a }}.\dfrac{{\sqrt a + 1}}{{a + 1}}\\
= \dfrac{{\left( {a – 1} \right)\left( {\sqrt a + 1} \right)}}{{\sqrt a \left( {a + 1} \right)}}\\
b)a = 3 + 2\sqrt 2 \left( {tmdk} \right)\\
= {\left( {\sqrt 2 + 1} \right)^2}\\
\Leftrightarrow \sqrt a = \sqrt 2 + 1\\
\Leftrightarrow A = \dfrac{{\left( {3 + 2\sqrt 2 – 1} \right)\left( {\sqrt 2 + 1 + 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {3 + 2\sqrt 2 + 1} \right)}}\\
= \dfrac{{\left( {2 + 2\sqrt 2 } \right)\left( {2 + \sqrt 2 } \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {4 + 2\sqrt 2 } \right)}}\\
= \dfrac{{2\left( {\sqrt 2 + 1} \right).\left( {2 + \sqrt 2 } \right)}}{{\left( {\sqrt 2 + 1} \right).2\left( {2 + \sqrt 2 } \right)}}\\
= 1\\
c)A < 0\\
\Leftrightarrow \dfrac{{\left( {a – 1} \right)\left( {\sqrt a + 1} \right)}}{{\sqrt a \left( {a + 1} \right)}} < 0\\
\Leftrightarrow a – 1 < 0\\
\Leftrightarrow a < 1\\
Vay\,0 < a < 1
\end{array}$