Question

$(4x+xy/ 4y+ y^{2} ) - (42 x^{2} zy^{2}/63zxy^{3} )$Alguem me explica por favor

Answer 1

[tex]$\left(\frac{4x+xy}{4y+ y^{2}}\right) - \left(\frac{42 x^{2} zy^{2}}{63zxy^{3}} \right)=$

\\

$

\left( \frac{[(4x+xy)\cdot(63zxy^3)]-[(4y+y^2)\cdot(42x^2zy^2)]}{ (4y+y^2)\cdot(63zxy^3)} \right)\\

\left( \frac{[252x^2y^3z+63x^2y^4z]-[168x^2y^3z+42x^2y^4z]}{(4y+y^2)\cdot(63xy^3z)}\right)\\

\left( \frac{252x^2y^3z+63x^2y^4z-168x^2y^3z-42x^2y^4z}{(4y+y^2)\cdot(63xy^3z)}\right)\\

\left( \frac{252x^2y^3z-168x^2y^3z+63x^2y^4z-42x^2y^4z}{(4y+y^2)\cdot(63xy^3z)}\right)\\

\left( \frac{(252-168)x^2y^3z+(63-42)x^2y^4z}{(4y+y^2)\cdot(63xy^3z)}\right)\\

\left( \frac{84x^2y^3z+21x^2y^4z}{(4y+y^2)\cdot(63xy^3z)}\right)\\

\left( \frac{(21x^2y^3z)\cdot(4+y^4)}{(4y+y^2)\cdot(63xy^3z)}\right)\\

\left( \frac{(\cancel{21}x^{\cancel{2}}\cancel{y^3}\cancel{z})\cdot(4+y^4)}{(4y+y^2)\cdot(\cancel{63}\cancel{x}\cancel{y^3}\cancel{z})}\right)\\

\left( \frac{x\cdot(4+y^4)}{(4y+y^2)\cdot3}\right)\\

\left( \frac{xy^4+4x}{3y^2+12y}\right)\\

$[\tex]