Cho C={1+ [√a/(a+1)]} ÷{[1/ (√a-1)]-[2√a/(a√a+√a-a-1)]}
a)rút gọn C
b)tính giá trị của a để C<1
c)tính C nếu a=19-8√3
d)tìm mọi a ∈ Z để C ∈ Z
Cho C={1+ [√a/(a+1)]} ÷{[1/ (√a-1)]-[2√a/(a√a+√a-a-1)]}
a)rút gọn C
b)tính giá trị của a để C<1
c)tính C nếu a=19-8√3
d)tìm mọi a ∈ Z để C ∈ Z
Đáp án:
$\begin{array}{l}
a)Dkxd:a \ge 0;a \ne 1\\
C = \left( {1 + \dfrac{{\sqrt a }}{{a + 1}}} \right):\left( {\dfrac{1}{{\sqrt a – 1}} – \dfrac{{2\sqrt a }}{{a\sqrt a + \sqrt a – a – 1}}} \right)\\
= \dfrac{{a + 1 + \sqrt a }}{{a + 1}}:\left( {\dfrac{1}{{\sqrt a – 1}} – \dfrac{{2\sqrt a }}{{\left( {a + 1} \right)\left( {\sqrt a – 1} \right)}}} \right)\\
= \dfrac{{a + \sqrt a + 1}}{{a + 1}}:\dfrac{{a + 1 – 2\sqrt a }}{{\left( {\sqrt a – 1} \right)\left( {a + 1} \right)}}\\
= \dfrac{{a + \sqrt a + 1}}{{a + 1}}.\dfrac{{\left( {\sqrt a – 1} \right)\left( {a + 1} \right)}}{{{{\left( {\sqrt a – 1} \right)}^2}}}\\
= \dfrac{{a + \sqrt a + 1}}{{\sqrt a – 1}}\\
b)C < 1\\
\Rightarrow \dfrac{{a + \sqrt a + 1}}{{\sqrt a – 1}} < 1\\
\Rightarrow \dfrac{{a + \sqrt a + 1}}{{\sqrt a – 1}} – 1 < 0\\
\Rightarrow \dfrac{{a + \sqrt a + 1 – \sqrt a + 1}}{{\sqrt a – 1}} < 0\\
\Rightarrow \dfrac{{a + 2}}{{\sqrt a – 1}} < 0\\
\Rightarrow \sqrt a – 1 < 0\\
\left( {do:a + 2 > 0\,khi:a \ge 0} \right)\\
\Rightarrow \sqrt a < 1\\
\Rightarrow a < 1\\
Vay\,0 \le a < 1\\
c)a = 19 – 8\sqrt 3 \left( {tmdk} \right)\\
= 16 – 2.4.\sqrt 3 + 3\\
= {\left( {4 – \sqrt 3 } \right)^2} = 4 – \sqrt 3 \\
\Rightarrow \sqrt a = 4 – \sqrt 3 \\
\Rightarrow C = \dfrac{{a + 2}}{{\sqrt a – 1}} = \dfrac{{19 – 8\sqrt 3 + 2}}{{4 – \sqrt 3 – 1}}\\
= \dfrac{{21 – 8\sqrt 3 }}{{3 – \sqrt 3 }} = \dfrac{{13 – \sqrt 3 }}{2}\\
d)C = \dfrac{{a + 2}}{{\sqrt a – 1}} = \dfrac{{a – \sqrt a + \sqrt a – 1 + 3}}{{\sqrt a – 1}}\\
= \dfrac{{\left( {\sqrt a – 1} \right)\left( {\sqrt a + 1} \right) + 3}}{{\sqrt a – 1}}\\
= \sqrt a + 1 + \dfrac{3}{{\sqrt a – 1}}\\
C \in Z\\
\Rightarrow \dfrac{3}{{\sqrt a – 1}} \in Z\\
\Rightarrow \left( {\sqrt a – 1} \right) \in \left\{ { – 1;1;3} \right\}\\
\Rightarrow \sqrt a \in \left\{ {0;2;4} \right\}\\
\Rightarrow a \in \left\{ {0;4;16} \right\}\left( {tmdk} \right)\\
Vay\,a \in \left\{ {0;4;16} \right\}
\end{array}$