a)3Fe+2O₂→Fe₃O₄.

b)2Fe+3Cl₂→2FeCl₃.

c)CH₄+2O₂→CO₂+2H₂O.

\(a,3Fe+2O_2\rightarrow^{t^o}Fe_3O_4\\ 3:2:1\\ b,2Fe+3Cl_2\rightarrow^{t^o}2FeCl_3\\ 2:3:2\\ c,CH_4+2O_2\rightarrow^{t^o}CO_2+2H_2O\\ 1:2:1:2\)

\(a,=2x\left(2x+3y\right)\\ b,=3x\left(x^2-9\right)=3x\left(x-3\right)\left(x+3\right)\\ c,=x\left(x-7y\right)+\left(x-7y\right)=\left(x+1\right)\left(x-7y\right)\\ d,=\left(x+y\right)^2-9z^2=\left(x+y-3z\right)\left(x+y+3z\right)\)

a. 5x + 10y = 5(x + 2y)

b. 3x³ - 12x

= 3x(x² - 4)

= 3x(x - 2)(x + 2)

c. 4x² + 9x - 4xy - 9y

= 4x² - 4xy + 9x - 9y

= 4x(x - y) + 9(x - y)

= (4x + 9)(x - y)

d. 3x² + 5y - 3xy - 5x

= 3x² - 3xy - 5x + 5y

= 3x(x - y) - 5(x - y)

= (3x - 5)(x - y)

e. 3y² - 3z² + 3x² + 6xy

= 3x² + 6xy + 3y² - 3z²

= 3(x² + 2xy + y² - z²)

= \(3\left[\left(x+y\right)^2-z^2\right]\)

= \(3\left(x+y+z\right)\left(x+y-z\right)\)

a) \(=5\left(x+2y\right)\)

b) \(=3x\left(x^2-4\right)=3x\left(x-2\right)\left(x+2\right)\)

c) \(=4x\left(x-y\right)+9\left(x-y\right)=\left(x-y\right)\left(4x+9\right)\)

d) \(=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

e) \(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]=3\left[\left(x+y\right)^2-z^2\right]\)

\(=3\left(x+y-z\right)\left(x+y+z\right)\)

\(a,=5\left(x+2y\right)\\ b,=3x\left(x^2-4\right)=3x\left(x-2\right)\left(x+2\right)\\ c,=x\left(4x+9\right)-y\left(4x+9\right)=\left(x-y\right)\left(4x+9\right)\\ d,=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

a, Vì ABCD là hbh nên AD//BC hay AM//BC và AD=AM=BC(tính chất đối xứng)

Do đó AMBC là hbh

b, Ta có AMBC là hbh

Mà N là giao điểm AB và MC nên N là trung điểm AB và MC

Vậy MN=NC

3-C

4-A

1) \(2\left(x-9\right)=12\Rightarrow x-9=12:2=6\)

\(\Rightarrow x=6+9=15\)

2) \(5x\left(x-3\right)-2\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{2}{5}\end{matrix}\right.\)

3) \(\left(x-4\right)\left(7x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{2}{7}\end{matrix}\right.\)

4) \(\left(x-1\right)\left(x+1\right)-x\left(x+3\right)=0\)

\(\Rightarrow x^2-1-x^2-3x=0\Rightarrow3x=-1\Rightarrow x=-\dfrac{1}{3}\)