bn ... ơi...mik ...bỏ...cuộc ...hu...hu

. Huhu T^T mong sẽ có ai đó giúp mình "((

mọi người giúp với 

b: \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{y}{3}\right)^2=\left(\dfrac{2}{3}\right)^6\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\\\left(\dfrac{y}{3}\right)^2=\left(\dfrac{8}{27}\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{2}{9}=\dfrac{4}{9}\\\dfrac{y}{3}=\dfrac{8}{27}\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{2}{9}=\dfrac{4}{9}\\\dfrac{y}{3}=-\dfrac{8}{27}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{8}{9}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{8}{9}\end{matrix}\right.\end{matrix}\right.\)

c: =>8x-1=5

=>8x=6

hay x=3/4

a: =>x-xy+y=0

=>x(1-y)+1-y-1=0

=>(x+1)(1-y)=1

=>(x+1)(y-1)=-1

=>\(\left(x+1;y-1\right)\in\left\{\left(-1;1\right);\left(1;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-2;2\right);\left(0;0\right)\right\}\)

b: 2x-xy-2y=3

=>x(2-y)-2y+4=7

=>x(2-y)+2(2-y)=7

=>(x+2)(y-2)=-7

=>\(\left(x+2;y-2\right)\in\left\{\left(1;-7\right);\left(-7;1\right);\left(-1;7\right);\left(7;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-1;-5\right);\left(-9;3\right);\left(-3;9\right);\left(5;1\right)\right\}\)

c: =>x(4-y)+5y-20=-3

=>x(4-y)-5(4-y)=-3

=>(4-y)(x-5)=-3

=>(x-5)(y-4)=3

=>\(\left(x-5;y-4\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(6;9\right);\left(8;5\right);\left(4;1\right);\left(2;3\right)\right\}\)