p=x^2-2x-3/ 2x^2-6x a) tìm đkxđ b) Rút gọn c) P=2/3 khi x=? d) x thuộc Z để P nhận giá trị nguyên 02/08/2021 Bởi Julia p=x^2-2x-3/ 2x^2-6x a) tìm đkxđ b) Rút gọn c) P=2/3 khi x=? d) x thuộc Z để P nhận giá trị nguyên
\(\begin{array}{l}a)DK:\,x \ne \left\{ {0;5} \right\}\\b)\\P = \frac{{{x^2} – 3x + x – 3}}{{2x\left( {x – 3} \right)}}\\ = \frac{{x\left( {x – 3} \right) + \left( {x – 3} \right)}}{{2x\left( {x – 3} \right)}}\\ = \frac{{\left( {x + 1} \right)\left( {x – 3} \right)}}{{2x\left( {x – 3} \right)}} = \frac{{x + 1}}{{2x}}\\c)\,P = \frac{2}{3} \Leftrightarrow \frac{{x + 1}}{{2x}} = \frac{2}{3}\\ \Rightarrow 3x + 3 = 6x\\ \Leftrightarrow x = 1\left( {tm} \right)\\d)\,P \in Z \Rightarrow 2P \in Z \Leftrightarrow \frac{{2x + 2}}{{2x}} = 1 + \frac{1}{x} \in Z\\ \Rightarrow x \in \left\{ { – 1;1} \right\}\\Voi\,x = 1 \Rightarrow P = 1\left( {tm} \right)\\Voi\,x = – 1 \Rightarrow P = 0\left( {tm} \right)\end{array}\) Bình luận
\(\begin{array}{l}
a)DK:\,x \ne \left\{ {0;5} \right\}\\
b)\\
P = \frac{{{x^2} – 3x + x – 3}}{{2x\left( {x – 3} \right)}}\\
= \frac{{x\left( {x – 3} \right) + \left( {x – 3} \right)}}{{2x\left( {x – 3} \right)}}\\
= \frac{{\left( {x + 1} \right)\left( {x – 3} \right)}}{{2x\left( {x – 3} \right)}} = \frac{{x + 1}}{{2x}}\\
c)\,P = \frac{2}{3} \Leftrightarrow \frac{{x + 1}}{{2x}} = \frac{2}{3}\\
\Rightarrow 3x + 3 = 6x\\
\Leftrightarrow x = 1\left( {tm} \right)\\
d)\,P \in Z \Rightarrow 2P \in Z \Leftrightarrow \frac{{2x + 2}}{{2x}} = 1 + \frac{1}{x} \in Z\\
\Rightarrow x \in \left\{ { – 1;1} \right\}\\
Voi\,x = 1 \Rightarrow P = 1\left( {tm} \right)\\
Voi\,x = – 1 \Rightarrow P = 0\left( {tm} \right)
\end{array}\)
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