Trắc nghiệm chuyên đề Toán 8 Chủ đề 10. Ôn tập và kiểm tra có đáp án (Đề kiểm tra 1 tiết - Đề 2)

Đáp án D

Đáp án A

Đáp án B

Đáp án C

Đáp án A

Đáp án C

$$\begin{gathered} \mathrm{a}) \frac{{\mathrm{x}}^2+2\mathrm{xy}+{\mathrm{y}}^2}{\mathrm{x}-\mathrm{y}}=\frac{\mathrm{A}}{{\mathrm{x}}^2-{\mathrm{y}}^2}\Rightarrow \frac{{(\mathrm{x}+\mathrm{y})}^2}{\mathrm{x}-\mathrm{y}}=\frac{\mathrm{A}}{{\mathrm{x}}^2-{\mathrm{y}}^2} \\ \Rightarrow \frac{{(\mathrm{x}+\mathrm{y})}^2(\mathrm{x}+\mathrm{y})}{(\mathrm{x}-\mathrm{y})(\mathrm{x}+\mathrm{y})}=\frac{\mathrm{A}}{{\mathrm{x}}^2-{\mathrm{y}}^2}\Rightarrow \frac{{(\mathrm{x}+\mathrm{y})}^3}{{\mathrm{x}}^2-{\mathrm{y}}^2}=\frac{\mathrm{A}}{{\mathrm{x}}^2-{\mathrm{y}}^2} \\ \mathrm{Vậy} \mathrm{A}={(\mathrm{x}+\mathrm{y})}^3={\mathrm{x}}^3+3{\mathrm{x}}^2\mathrm{y}+3{\mathrm{xy}}^2+{\mathrm{y}}^3 \end{gathered}$$

$$\begin{gathered} \mathrm{b}) \frac{5{\mathrm{x}}^2-125}{{\mathrm{x}}^2+1}=0\Rightarrow 5{\mathrm{x}}^2-125=0 (\mathrm{vì} {\mathrm{x}}^2+1>0) \\ \iff 5({\mathrm{x}}^2-25)=0 \iff 5(\mathrm{x}+5\left)\right(\mathrm{x}-5)=0 \\ \iff [\begin{array}{c}\mathrm{x}+5=0\\ \mathrm{x}-5=0\end{array}\iff [\begin{array}{c}\mathrm{x}=5\\ \mathrm{x}=-5\end{array} \\ \mathrm{Vậy} \mathrm{x}=-5, \mathrm{x}=5 \end{gathered}$$

$$\begin{gathered} \mathrm{b}) \frac{{\mathrm{x}}^3}{2\mathrm{y}(\mathrm{x}-2\mathrm{y})}:\frac{{\mathrm{x}}^2}{2{\mathrm{x}}^2\mathrm{y}-8{\mathrm{y}}^3} \\ =\frac{{\mathrm{x}}^3}{2\mathrm{y}(\mathrm{x}-2\mathrm{y})}.\frac{2\mathrm{y}({\mathrm{x}}^2-4{\mathrm{y}}^2)}{{\mathrm{x}}^2} \\ =\frac{\mathrm{x}(\mathrm{x}+2\mathrm{y})(\mathrm{x}-2\mathrm{y})}{\mathrm{x}-2\mathrm{y}} \\ =(\mathrm{x}+2\mathrm{y})\mathrm{x} \end{gathered}$$

$$\begin{gathered} \mathrm{a}) {\mathrm{x}}^2-9\ne 0\Rightarrow (\mathrm{x}+3\left)\right(\mathrm{x}-3)\ne 0 \\ \Rightarrow \{\begin{array}{c}\mathrm{x}+3\ne 0\\ \mathrm{x}-3\ne 0\end{array}\Rightarrow \{\begin{array}{c}\mathrm{x}\ne -3\\ \mathrm{x}\ne 3\end{array} \\ \mathrm{Vậy} \mathrm{A}=\frac{{\mathrm{x}}^2-6\mathrm{x}+9}{{\mathrm{x}}^2-9} \mathrm{xác} \mathrm{định} \mathrm{khi} \mathrm{x}\ne 3 \mathrm{và} \mathrm{x}\ne -3 \end{gathered}$$

$$\begin{gathered} \mathrm{b}) \mathrm{Với} \mathrm{x}\ne -3 \mathrm{và} \mathrm{x}\ne 3, \mathrm{ta} \mathrm{có} \\ \mathrm{A}=\frac{{\mathrm{x}}^2-6\mathrm{x}+9}{{\mathrm{x}}^2-9}=\frac{{(\mathrm{x}-3)}^2}{(\mathrm{x}+3)(\mathrm{x}-3)}=\frac{\mathrm{x}-3}{\mathrm{x}+3} \end{gathered}$$

$$\begin{gathered} \mathrm{c}) \mathrm{A}=0 \iff \left\{\begin{array}{c}{\mathrm{x}}^2-6\mathrm{x}+9=0\\ {\mathrm{x}}^2-9\ne 0\end{array}\right.\iff \left\{\begin{array}{c}{(\mathrm{x}-3)}^2=0\\ \mathrm{x}\ne 3; \mathrm{x}\ne -3\end{array}\right. \\ \left\{\begin{array}{c}\mathrm{x}-3=0\\ \mathrm{x}\ne 3; \mathrm{x}\ne -3\end{array}\right.\iff \left\{\begin{array}{c}\mathrm{x}=3\\ \mathrm{x}\ne 3; \mathrm{x}\ne -3\end{array}\right.(\mathrm{loại}) \\ \mathrm{Vậy} \mathrm{không} \mathrm{có} \mathrm{giá} \mathrm{trị} \mathrm{nào} \mathrm{của} \mathrm{x} \mathrm{để} \mathrm{A}=0 \end{gathered}$$