Question

calculand (- x^2/y^4): (-x^4/y^2) unde x si y sunt numere reale diferite de 0 vei obtine

Answer 1

 

[tex]\displaystyle\\ (-):(-) = +\\ \Rightarrow~~\text{semnele minus vor disparea}\\\\ (- x^2/y^4): (-x^4/y^2)=\\\\ =\left(-\frac{x^2}{y^4} \right):\left(-\frac{x^4}{y^2} \right)=\frac{x^2}{y^4}:\frac{x^4}{y^2} =\frac{x^2}{y^4}\cdot\frac{y^2}{x^4} =\boxed{\frac{1}{x^2y^2}}[/tex]

Answer 2

$( - \frac{ {x}^{2} }{ {y}^{4} }) : ( - \frac{ {x}^{4} }{ {y}^{2} }) = \frac{ - {x}^{2} }{ {y}^{4} } : - \frac{ {x}^{4} }{ {y}^{2} } = - \frac{ {x}^{2} }{ {y}^{4} } : - \frac{ {x}^{4} }{ {y}^{2} } = \frac{ {x}^{2} }{ {y}^{4} } : \frac{ {x}^{4} }{ {y}^{2} } = \frac{ {x}^{2} }{ {y}^{4} } \times \frac{ {y}^{2} }{ {x}^{4} } = \frac{ {x}^{2} {y}^{2} }{ {y}^{4} {x}^{4} } = {x}^{2 - 4}y ^{2 - 4} = {x}^{ - 2} {y}^{ - 2} = \frac{1}{ {x}^{2} } {y}^{ - 2} = \frac{1}{ {x}^{2} } \times \frac{1}{ {y}^{2} } = \frac{1}{ {x}^{2} {y}^{2} }$