Bài tập theo tuần Toán 8 - Tuần 10 (Đề 2)

Đáp án C

Đáp án D

Đáp án D

Đáp án B

1 - S

2 - S

3 - S

4 - Đ

$\begin{array}{l}\left(12{\mathrm{x}}^3{\mathrm{y}}^3-3{\mathrm{x}}^2{\mathrm{y}}^3+4{\mathrm{x}}^2{\mathrm{y}}^4\right):6{\mathrm{x}}^2{\mathrm{y}}^3\\ =\left(12{\mathrm{x}}^3{\mathrm{y}}^3:6{\mathrm{x}}^2{\mathrm{y}}^3\right)-\left(3{\mathrm{x}}^2{\mathrm{y}}^3:6{\mathrm{x}}^2{\mathrm{y}}^3\right)+\left(4{\mathrm{x}}^2{\mathrm{y}}^4:6{\mathrm{x}}^2{\mathrm{y}}^3\right)\\ =2\mathrm{x}-\frac{1}{2}+\frac{2}{3}\mathrm{y}\end{array}$

$\begin{array}{l}{135}^2+{20}^2-{15}^2-40.135\\ =\left({135}^2-\mathrm{2.135.20}+{20}^2\right)-{15}^2\\ ={\left(135-20\right)}^2-{15}^2\\ ={115}^2-{15}^2=\left(115+15\right)\left(115-15\right)\\ =130.100=13000\end{array}$

$2\mathrm{x}\left(\mathrm{x}+3\right)+2\left(\mathrm{x}+3\right)=2\left(\mathrm{x}+1\right)\left(\mathrm{x}+3\right)$

$\begin{array}{l}{\mathrm{x}}^3-3{\mathrm{x}}^2-4\mathrm{x}+12={\mathrm{x}}^2\left(\mathrm{x}-3\right)-4\left(\mathrm{x}-3\right)\\ =\left(\mathrm{x}-3\right)\left({\mathrm{x}}^2-4\right)=\left(\mathrm{x}-3\right)\left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)\end{array}$

$\begin{array}{l}{\mathrm{x}}^2\left(\mathrm{x}+3\right)-4\mathrm{x}-12=0\\ {\mathrm{x}}^2\left(\mathrm{x}+3\right)-4\left(\mathrm{x}+3\right)=0\\ \left(\mathrm{x}+3\right)\left({\mathrm{x}}^2-4\right)=0\\ \left(\mathrm{x}+3\right)\left(\mathrm{x}+2\right)\left(\mathrm{x}-2\right)=0\Rightarrow \left[\begin{array}{l}\mathrm{x}=-3\\ \mathrm{x}=-2\\ \mathrm{x}=2\end{array}\right.\end{array}$