a) \(=\dfrac{x+15}{\left(x-3\right)\left(x+3\right)}+\dfrac{2}{x+3}=\dfrac{x+15+2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

b) \(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{y^2+x^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x+y\right)^2-\left(x-y\right)^2+2\left(x^2+y^2\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2\left(x^2+y^2+2xy\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}\)

Ta có:

\(B=4x\left(2x+y\right)+2y\left(2x+y\right)-y\left(y+2x\right)\)

\(\Leftrightarrow B=\left(4x+2y-y\right)\left(2x+y\right)=\left(4x+y\right)\left(2x+y\right)=\left(4.\dfrac{1}{2}+\dfrac{-3}{5}\right)\left(2.\dfrac{1}{2}+\dfrac{-3}{5}\right)=\dfrac{14}{25}\)

a. Ta có :

x²-9= (x-3)(x+3)

=> Mtc : (x-3)(x+3)

Nhân tử phụ 1 : (x-3)(x+3):(x-3)(x+3)=1

Nhân tử phụ 2 : (x-3)(x+3):(x+3)=x-3

Qui đồng:

x+15/x²-9= x+15/(x-3)(x+3)=(x-15).1/(x-3)(x+3).1=x-15/(x-3)(x+3)

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

        x-15/(x-3)(x+3)       +      2(x-3)/(x-3)(x+3)

  =   (x-15)+2(x-3)/(x-3)(x+3)

  =    x-15+2/x+3 

  =  x-13/x+3

\(a,=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}\cdot\dfrac{2}{-\left(x-3\right)}=\dfrac{x-3}{2}\cdot\dfrac{2}{-\left(x-3\right)}=-1\\ b,=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{\left(x-y\right)}=2\)

a,\(\dfrac{x^2-9}{2x+6}:\dfrac{3-x}{2}=\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}.\dfrac{2}{3-x}=\dfrac{x-3}{3-x}=\dfrac{-\left(3-x\right)}{3-x}=-1\)

b, \(\dfrac{2x}{x-y}-\dfrac{2y}{x-y}=\dfrac{2x-2y}{x-y}=\dfrac{2\left(x-y\right)}{x-y}=2\)

a, \(=5x^3-2x^2y-5x^2y+2xy^2+5x-2y\)

\(=5x^3-7x^2y+2xy^2+5x-2y\)

b, \(=\left(x^2-1\right)\left(x+2\right)\)

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

c, đề không rõ

d, đề không rõ

P/s có gì bạn tham khảo các thanh công cụ ở trên để đánh cô hỏi cho rõ nha

\(a)\left(x+3y\right)\left(x-2y\right)\\ =x^3-2xy+3xy-6y^2\\ =x^2+xy-6y^2\\ b)\left(2x-y\right)\left(y-5x\right)\\ = 2xy-10x^2-y^2+5xy\\ =7xy-10x^2-y^2\\ c)\left(2x-5y\right)\left(y^2-2xy\right)\\ =2xy^2-4x^2y-5y^3+10xy^2\\ =12xy^2-4x^2y-5y^2\\ d)\left(x-y\right)\left(x^2-xy-y^2\right)\\ =x^3-x^2y-xy^2-x^2y+xy^2+y^3\\ =x^3-2x^2y+y^3\)

Giúp với mọi người

c: \(=3x^3-4x^2+6x^2-8x=3x^2+2x^2-8x\)

\(a,=27x^3+27x^2+9x+1\)

\(b,=\dfrac{x^3}{27}-\dfrac{x^2}{3}+x-1\)

\(c,=-\left(27x^3-27x^2y^2+9xy^4-y^6\right)\)

\(=-27x^3+27x^2y^2-9xy^4+y^6\)

\(d,=\dfrac{x^3}{y^3}-\dfrac{6x}{y}+\dfrac{12y}{x}-\dfrac{8y^3}{x^3}\)

a) \(\left(3x+1\right)^3=27x^3+27x^2+9x+1\)

b) \(\left(\dfrac{x}{3}-1\right)^3=\dfrac{x^3}{27}-\dfrac{x^2}{3}\)

c) \(\left(-y^2+3x\right)^3=27x^3-27x^2y^2+9xy^4-y^6\)

d) \(\left(\dfrac{x}{y}-\dfrac{2y}{x}\right)^3=\dfrac{x^3}{y^3}-\dfrac{6x}{y}+\dfrac{12y}{x}-\dfrac{8y^3}{x^3}\)

a: \(=x-\dfrac{3}{2}+2y\)

b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)