a) Ta có : \(\left|5x+15\right|\ge0\) với mọi \(x\)

\(\left|3y-2y\right|\ge0\) với mọi \(x,y\)

Nên \(\left|5x+15\right|+\left|3y-2x\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}5x+15=0\\3y-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\3y-2\times\left(-3\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\)

Vậy........

Học tốt!!

Ta có:x+6y+3xy+35=0

  <=> x(1+3y)+2(1+3y)+33=0

  <=> (1+3y)(x+2)=-33

Do \(y\in Z\Rightarrow3y⋮3\Rightarrow1+3y\) chia 3 dư 1

Mà trong các Ư_(-33) chỉ có -11 và 1 là chia 3 dư 1

TH1.\(\left\{{}\begin{matrix}1+3y=1\\x+2=-33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-35\end{matrix}\right.\)

TH2.\(\left\{{}\begin{matrix}1+3y=-11\\x+2=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-4\\x=1\end{matrix}\right.\)

Vậy,(x;y)=(-35;0),(1;-4)

\(a,=2\left(xy^2-2\right)\\ b,=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x-y+1\right)\left(x+y\right)\\ c,=y\left(x^2-6x-9\right)\\ d,=x^2+3x+x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\)

\(a,-x^3+\left(x-3\right)\left[\left(2x+1\right)^2-2\left(\dfrac{3}{2}x^2+\dfrac{1}{2}x-4\right)\right]\\ =-x^3+\left(x-3\right)\left(4x^2+4x+1-3x^2-x+8\right)\\ =-x^3+\left(x-3\right)\left(x^2+3x+9\right)\\ =-x^3+\left(x^3-27\right)=-27\)

\(b,\left(x+2y\right)^3-\left(x-3y\right)\left(x^2+3xy+9y^2\right)-6y\left(x^2+2xy-\dfrac{35}{6}y^2\right)\\ =x^3+6x^2y+12xy^2+8y^3-x^3+27y^3-6x^2y-12xy^2+35y^3\\ =0\)