cho các biểu thức sau
B = $\frac{2}{x+\sqrt{x}+1}$ ; C = $\frac{5}{x+\sqrt{x}+2 }$ ; D = $\frac{\sqrt{x}+6 }{\sqrt{x}+1 }$
tìm x để B, C, D nhận được giá trị nguyên?
cho các biểu thức sau B = $\frac{2}{x+\sqrt{x}+1}$ ; C = $\frac{5}{x+\sqrt{x}+2 }$ ; D = $\frac{\sqrt{x}+6 }{\sqrt{x}+1 }$ tìm x để B, C, D nhận đ
By Amara
Đáp án:
$\begin{array}{l}
B = \dfrac{2}{{x + \sqrt x + 1}} \in Z\\
\Rightarrow x + \sqrt x + 1 \in U\left( 2 \right) = \left\{ { – 2; – 1;1;2} \right\}\\
Do:x + \sqrt x + 1\\
= x + 2.\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge 1\\
\Rightarrow \left[ \begin{array}{l}
x + \sqrt x + 1 = 1\\
x + \sqrt x + 1 = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x + \sqrt x = 0\\
x + \sqrt x – 1 = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x .\left( {\sqrt x + 1} \right) = 0\\
{\left( {\sqrt x + \dfrac{1}{2}} \right)^2} = \dfrac{5}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
\sqrt x + \dfrac{1}{2} = \dfrac{{\sqrt 5 }}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
\sqrt x = \dfrac{{\sqrt 5 – 1}}{2}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{{3 – \sqrt 5 }}{2}
\end{array} \right.\\
C = \dfrac{5}{{x + \sqrt x + 2}} \in Z\\
Do:x + \sqrt x + 2 \ge 2\\
\Rightarrow \left( {x + \sqrt x + 2} \right) = 5\\
\Rightarrow x + \sqrt x – 3 = 0\\
\Rightarrow x + 2.\dfrac{1}{2}.\sqrt x + \dfrac{1}{4} = \dfrac{{13}}{4}\\
\Rightarrow {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} = \dfrac{{13}}{4}\\
\Rightarrow \sqrt x = \dfrac{{\sqrt {13} – 1}}{2}\\
\Rightarrow x = \dfrac{{7 – \sqrt {13} }}{2}\\
D = \dfrac{{\sqrt x + 6}}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 1 + 5}}{{\sqrt x + 1}}\\
= 1 + \dfrac{5}{{\sqrt x + 1}}\\
D \in Z\\
\Rightarrow \dfrac{5}{{\sqrt x + 1}} \in Z\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x + 1 = 1\\
\sqrt x + 1 = 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 16
\end{array} \right.
\end{array}$