Câu 10:
\(\begin{array}{l}a)\,\,2{x^2} - 6xy = 2x\left( {x - 3y} \right).\\b)\,\,{x^2} - 2xy + x - 2y = x\left( {x - 2y} \right) + \left( {x - 2y} \right)\\ = \left( {x - 2y} \right)\left( {x + 1} \right).\\c)\,\,{x^2} + 4{y^2} + 4xy - 16\\ = {x^2} + 4xy + 4{y^2} - 16\\ = {\left( {x + 2y} \right)^2} - {4^2}\\ = \left( {x + 2y - 4} \right)\left( {x + 2y + 4} \right).\end{array}\)
Câu 11:
\(A = \left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}}} \right).\frac{{{x^2} - 4x + 4}}{4}\)
a) Điều kiện: \(\left\{ \begin{array}{l}x \ne 2\\x \ne - 2\end{array} \right..\)
\(\begin{array}{l}A = \left( {\frac{1}{{x - 2}} - \frac{1}{{x + 2}}} \right).\frac{{{x^2} - 4x + 4}}{4}\\ = \frac{{x + 2 - x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\frac{{{{\left( {x + 2} \right)}^2}}}{4}\\ = \frac{4}{{x - 2}}.\frac{{x + 2}}{4} = \frac{{x + 2}}{{x - 2}}.\end{array}\)
b) Điều kiện: \(x \ne \pm 2.\)
Ta có:\(A = \frac{{x + 2}}{{x - 2}} = \frac{{x - 2 + 4}}{{x - 2}} = 1 + \frac{4}{{x - 2}}.\)
\( \Rightarrow A \in \mathbb{Z} \Leftrightarrow \frac{4}{{x - 2}} \in \mathbb{Z} \Leftrightarrow 4\,\, \vdots \,\,\left( {x - 2} \right)\)
\(\begin{array}{l} \Rightarrow x - 2 \in U\left( 4 \right) = \left\{ { \pm 1; \pm 2; \pm 4} \right\}\\ \Rightarrow \left[ \begin{array}{l}x - 2 = 1\\x - 2 = - 1\\x - 2 = - 2\\x - 2 = 2\\x - 2 = - 4\\x - 2 = 4\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3\,\,\,\left( {tm} \right)\\x = 1\,\,\,\left( {tm} \right)\\x = 0\,\,\,\left( {tm} \right)\\x = 4\,\,\,\left( {tm} \right)\\x = - 2\,\,\,\left( {ktm} \right)\\x = 6\,\,\,\left( {tm} \right)\end{array} \right..\end{array}\)
Vậy \(x \in \left\{ {3;\,\,1;\,\,0;\,\,4;\,\,6} \right\}.\)
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