a: \(=\dfrac{3x+6-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+10}{x^2-4}\)

b: \(=\dfrac{10x+15+4x-6+2x+5}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x+14}{\left(2x-3\right)\left(2x+3\right)}\)

a: =>3M+2x^4y^4=x^4y^4

=>3M=-x^4y^4

=>M=-1/3*x^4y^4

b: x^2-2M=3x^2

=>2M=-2x^2

=>M=-x^2

c: =>M=-x^2y^3-3x^2y^3=-4x^2y^3

d: =>M=7x^2y^2-3x^2y^2=4x^2y^2

a: =1/3x^3-2x^2-7/3x

b: =5/4xyz^2+1/4xyz

c: =9/8x^2y^3-3x^2y

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người dễ theo dõi hơn.

1) \(3x^2-9=3\left(x^2-3\right)\)

2) \(\dfrac{1}{2}x^2-2y^2\)

\(=\dfrac{1}{2}\left(x^2-4y^2\right)\)

\(=\dfrac{1}{2}\left(x-2y\right)\left(x +2y\right)\)

3) \(3x^2-12y^2\)

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

\(=3\left(x-2y\right)\left(x+2y\right)\)

4) \(\dfrac{1}{3}x^2y^2-3x^2\)

\(=\dfrac{1}{3}x^2\left(y^2-9\right)\)

\(=\dfrac{1}{3}x^2\left(y-3\right)\left(y+3\right)\)

#Toru

a/ (x-1)²-(4x+3)(2-x)=x²-2x+1-(8x-4x²+6-3x)

=x²-2x+1-8x+4x²-6+3x=5x²-7x-6

b/ (15x³y² - 6x²y³) : 3x²y² = 5x - 2y

c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)

a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2

b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y

=>A-B=12xy^2-14x^2y

c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2

=>A-B=-5x^2y^3-x^3y^2

d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2

\(Q=x^2+2xy+\left(-3x^3+3x^3\right)+\left(2y^3-y^3\right)=x^2+2xy+y^3\)

\(P=\left(\dfrac{1}{3}x^2y-\dfrac{1}{3}x^2y\right)+\left(xy^2+\dfrac{1}{2}xy^2\right)-\left(xy+5xy\right)=\dfrac{3}{2}xy^2-6xy\)