Question

[Ngô Thị Phương Thảo]

Chia đa thức cho đơn thức

a) \(\left(-2\right)^5:\left(-2\right)^3\)
b) \(\left(-y\right)^7:\left(-y\right)^3\)
c) \(x^{12}:\left(-x^{10}\right)\)
d) \(\left(2x^6\right):\left(2x\right)^3\)
e) \(\left(-3x\right)^5:\left(-3x\right)^2\)
f) \(\left(xy^2\right)^4:\left(xy^2\right)^2\)
i) \(\left(x+2\right)^9:\left(x+2\right)^6\)

Answer 1

[Ngô Thanh Sang]

a) \(A=\dfrac{\left(-2\right)^5}{\left(-2\right)^3}=\left(-2\right)^{5-3}=\left(-2\right)^2=4\)

b) \(y\ne0:B=\dfrac{\left(-y\right)^7}{\left(-y\right)^3}=\left(-y\right)^{7-3}=\left(-y\right)^4=y^4\)

c) \(x\ne0:C=\dfrac{\left(x\right)^{12}}{\left(-x\right)^{10}}=\left(x\right)^{12-10}=\left(x\right)^2=x^4\)

d) \(x\ne0:D=\dfrac{2x^6}{\left(2x\right)^3}=\dfrac{2x^6}{8x^3}=\dfrac{1}{4}\left(x\right)^{6-3}=\dfrac{1}{4}\left(x\right)^3\)

e) \(x\ne0:E=\dfrac{\left(-3x\right)^5}{\left(-3x\right)^2}=\left(-3x\right)^{5-2}=\left(-3x\right)^3=-27x^3\)

f) \(x,y\ne0:F=\dfrac{\left(xy^2\right)^4}{\left(xy^2\right)^2}=\left(xy^2\right)^{4-2}=\left(xy^2\right)^2=x^2y^4\)

i) \(x\ne-2:I=\dfrac{\left(x+2\right)^9}{\left(x+2\right)^6}=\left(x+2\right)^{9-6}=\left(x+2\right)^3\)

Answer 2

[Trần Ngọc Bích]

A),(-2)⁵:(-2)³=(-2)²=4

B) (-y)⁷ :(-y)³=y⁴