Tìm x thuộc z để giá trị các biểu thức sau là số nguyên:
a. 5/x+3 b. x+5/x+3 c. 2x/x+3
d. x-2/ x+3 e. x/2x+1
f. 5/ căn x +1 g. căn x +3/ x+1 h. căn x -2/ căn x+1
Tìm x thuộc z để giá trị các biểu thức sau là số nguyên:
a. 5/x+3 b. x+5/x+3 c. 2x/x+3
d. x-2/ x+3 e. x/2x+1
f. 5/ căn x +1 g. căn x +3/ x+1 h. căn x -2/ căn x+1
Đáp án:
h. \(\left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne – 3\\
\dfrac{5}{{x + 3}} \in Z\\
\Leftrightarrow x + 3 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
x + 3 = 5\\
x + 3 = – 5\\
x + 3 = 1\\
x + 3 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = – 8\\
x = – 2\\
x = – 4
\end{array} \right.\\
b.DK:x \ne – 3\\
B = \dfrac{{x + 5}}{{x + 3}} = \dfrac{{x + 3 + 2}}{{x + 3}} = 1 + \dfrac{2}{{x + 3}}\\
B \in Z\\
\to \dfrac{2}{{x + 3}} \in Z\\
\to x + 3 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 3 = 2\\
x + 3 = – 2\\
x + 3 = 1\\
x + 3 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = – 1\\
x = – 5\\
x = – 2\\
x = – 4
\end{array} \right.\\
c.DK:x \ne – 3\\
C = \dfrac{{2x}}{{x + 3}} = \dfrac{{2\left( {x + 3} \right) – 6}}{{x + 3}}\\
= 2 – \dfrac{6}{{x + 3}}\\
C \in Z\\
\Leftrightarrow \dfrac{6}{{x + 3}} \in Z\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 6\\
x + 3 = – 6\\
x + 3 = 3\\
x + 3 = – 3\\
x + 3 = 2\\
x + 3 = – 2\\
x + 3 = 1\\
x + 3 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = – 9\\
x = 0\\
x = – 6\\
x = – 1\\
x = – 5\\
x = – 2\\
x = – 4
\end{array} \right.\\
d.D = \dfrac{{x – 2}}{{x + 3}} = \dfrac{{x + 3 – 5}}{{x + 3}} = 1 – \dfrac{5}{{x + 3}}\\
D \in Z\\
\Leftrightarrow \dfrac{5}{{x + 3}} \in Z\\
\Leftrightarrow x + 3 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
x + 3 = 5\\
x + 3 = – 5\\
x + 3 = 1\\
x + 3 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = – 8\\
x = – 2\\
x = – 4
\end{array} \right.\\
f.DK:x \ge 0\\
\dfrac{5}{{\sqrt x + 1}} \in Z\\
\to \sqrt x + 1 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 5\\
\sqrt x + 1 = – 5\\
\sqrt x + 1 = 1\\
\sqrt x + 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 4\\
\sqrt x = – 6\left( l \right)\\
\sqrt x = 0\\
\sqrt x = – 2\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 16\\
x = 0
\end{array} \right.\\
g.G = \dfrac{{\sqrt x + 3}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 + 2}}{{\sqrt x + 1}} = 1 + \dfrac{2}{{\sqrt x + 1}}\\
G \in Z\\
\to \dfrac{2}{{\sqrt x + 1}} \in Z\\
\to \sqrt x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 2\\
\sqrt x + 1 = – 2\\
\sqrt x + 1 = 1\\
\sqrt x + 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 1\\
\sqrt x = – 3\left( l \right)\\
\sqrt x = 0\\
\sqrt x = – 2\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 0
\end{array} \right.\\
h.H = \dfrac{{\sqrt x – 2}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 – 3}}{{\sqrt x + 1}} = 1 – \dfrac{3}{{\sqrt x + 1}}\\
H \in Z\\
\Leftrightarrow \dfrac{3}{{\sqrt x + 1}} \in Z\\
\Leftrightarrow \sqrt x + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 3\\
\sqrt x + 1 = – 3\left( l \right)\\
\sqrt x + 1 = 1\\
\sqrt x + 1 = – 1\left( l \right)
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.
\end{array}\)