Bài tập theo tuần Toán 8 - Tuần 11 (đề 1)

Đáp án D

Đáp án A

Đáp án B

Đáp án B

a - 5

b - 2

c - 4

d - 3

$\begin{array}{l}\left(6\mathrm{yz}-3\mathrm{xy}+20\mathrm{yz}\right):2\mathrm{y}\\ =6\mathrm{yz}:2\mathrm{y}-3\mathrm{xy}:2\mathrm{y}+20\mathrm{yz}:2\mathrm{y}\\ =3\mathrm{z}-\frac{3}{2}\mathrm{x}+10\mathrm{z}\end{array}$

$46.54=\left(50-4\right)\left(50+4\right)={50}^2-4^2=2500-16=2484$

${\mathrm{y}}^2+4{\mathrm{x}}^2+4\mathrm{xy}={\mathrm{y}}^2+2\mathrm{y}.2\mathrm{x}+{\left(2\mathrm{x}\right)}^2={\left(\mathrm{y}+2\mathrm{x}\right)}^2$

$4x-20y=4\left(x-5y\right)$

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

$\mathrm{x}\left(2\mathrm{x}+1\right)=0\iff \left[\begin{array}{l}\mathrm{x}=0\\ \mathrm{x}=-\frac{1}{2}\end{array}\right.$

$\begin{array}{l}{\left(\mathrm{x}-1\right)}^2=5\mathrm{x}\left(\mathrm{x}-1\right)\\ \iff \left(\mathrm{x}-1\right)\left(\mathrm{x}-1-5\mathrm{x}\right)=0\\ \iff \left(\mathrm{x}-1\right)\left(-4\mathrm{x}-1\right)=0\Rightarrow \left[\begin{array}{l}\mathrm{x}=1\\ \mathrm{x}=-\frac{1}{4}\end{array}\right.\end{array}$