Question

giúp mk mình cần gấp lắm

a,\(\dfrac{x^2+y^2-xy}{x^2-y^2}:\dfrac{x^3+y^3}{x^2+y^2-2xy}\)

b,\(\dfrac{x^3y+xy^3}{x^4y}:\left(x^2+y^2\right)\)

c,\(\dfrac{x^2-xy}{y}:\dfrac{x^2-xy}{xy+y}:\dfrac{x^2-1}{x^2+y}\)

d,\(\dfrac{x^2+y}{y}:\left(\dfrac{z}{x^2}:\dfrac{xy}{x^2y}\right)\)

e,\(\dfrac{x^2+1}{x}:\dfrac{x^2+1}{x-1}:\dfrac{x^3-1}{x^2+x}:\dfrac{x^2+2x+1}{x^2+x+1}\)

g,\(\left(\dfrac{z}{x^2}:\dfrac{xy}{x^2y}\right)\dfrac{x^2+y}{y}\)

Answer 1

a,\(\frac{x^2+y^2-xy}{x^2-y^2}:\frac{x^3+y^3}{x^2+y^2-2xy} =\frac{x^2+y^2-xy}{(x-y)(x+y)}\frac{(x+y)^2}{(x+y) (x^2-xy+y^2)}=\frac{1}{x-y} \)

b,\(\frac{x^3y+xy^3}{x^4y}:(x^2+y^2)=\frac{xy(x^2+y^2)}{x^4y(x^2+y^2)}=\frac{1}{x^3} \)

c,\(\frac{x^2-xy}{y}:\frac{x^2-xy}{xy+y}:\frac{x^2-1}{x^2+y} =\frac{x(x-y)y(x+y)(x^2+y)}{yx(x-y)(x^2-1)} =\frac{(x^2+y)(x+y)}{x^2-1} \)

d,\(\frac{x^2+y}{y}:(\frac{z}{x^2}:\frac{xy}{x^2y})=\frac{x^2+y}{ y}:(\frac{z}{x^2}\frac{x^2y}{xy})=\frac{x^2+y}{y}\frac{z}{x} \)