Trắc nghiệm Phân tích đa thức thành nhân tử Toán 8

$x\left(\text{xy}^2\text{z + y}^2z^2\text{+ xyz}^2\right)$

$y\left(x^2\text{yz + xyz}^2\text{+ x}^2z^2\right)$

$z\left(x^2y^2\text{+ xy}^2\text{z + x}^2\text{yz}\right)$

$\text{xyz}\left(\text{xy + yz + xz}\right)$

$4a^2-b$

$4a^2+7b$

$4a^2-7b$

$4a^2\text{+ b}$

$\left(x+3\right)\left(x-3\right)$

$\left(x-1\right)\left(x+9\right)$

$\left(x+3\right)^2$

$\left(x+6\right)\left(x-3\right)$

$\left(x+1\right)\left(x-y\right)$

$\left(x-y\right)\left(x-1\right)$

$\left(x-y\right)\left(x+y\right)$

$x\left(x-y\right)$

$\left(x-1\right)^3+2\left(x-1\right)^2=\left(x-1\right)^2\left(x+1\right)$

$\left(x-1\right)^3+2\left(x-1\right)=\left(x-1\right)\left[\left(x-1\right)^2+2\right]$

$\left(x-1\right)^3+2\left(x-1\right)^2=\left(x-1\right)\left[\left(x-1\right)^2+2x-2\right]$

$\left(x-1\right)^3+2\left(x-1\right)^2=\left(x-1\right)\left(x+3\right)$

x + 2

3(x – 2)

$\left(x-2\right)^2$

$\left(x+2\right)^2$

$\left(x-y-3\right)\left(x-\text{y + 3}\right)$

$\left(x-y-9\right)\left(x-\text{y + 9}\right)$

$\left(x+y-3\right)\left(x+\text{y + 3}\right)$

$\left(x+y-9\right)\left(x+y-9\right)$

$x^2+1$

$\left(x+1\right)^2$

$x^2-1$

$x^2+x+1$

0

1

2

3

x = 1

x = – 1

x = 2

x = 5

$\left(3x-2\right)\left(x+3\right)\left(x-5\right)$

$3\left(x-2\right)\left(x+3\right)\left(x-5\right)$

$\left(3x-2\right)\left(x-3\right)\left(x+5\right)$

$\left(x-2\right)\left(\text{3x}+3\right)\left(x-5\right)$

m > − 59

m < 0

$m⋮9$

m là số nguyên tố

B < 8 300

B > 8 500

B < 0

B > 8 300