Respuesta :

Hrishii

Answer:

[tex]\bigg(\frac{2}{3} {y} \bigg)^{2}[/tex]

STEP BY STEP EXPLANATION

[tex] \frac{1}{4} {x}^{2} - \bigg( \frac{2}{3} \bigg)xy \\ \\ = \bigg(\frac{1}{2} {x} \bigg)^{2} - 2. \bigg(\frac{1}{2} {x} \bigg)\bigg( \frac{2}{3} y \bigg ) + \bigg(\frac{2}{3} {y} \bigg)^{2} \\ \\ = \bigg(\frac{1}{2} {x} - \frac{2}{3} y\bigg)^{2} \\ [/tex]

To make [tex] \red{\bold{\frac{1}{4} {x}^{2} - \bigg( \frac{2}{3} \bigg)xy}} [/tex] a perfect square we should add [tex]\purple{\bold{\bigg(\frac{2}{3} {y} \bigg)^{2}}}[/tex]

mhanifa

Answer:

  • [tex]4/9y^2[/tex]

Step-by-step explanation:

Given:

  • [tex](1/4)x^2-(2/3)xy[/tex]

Consider the identity:

  • [tex](a - b)^2=a^2-2ab+b^2[/tex]

We can see that:

  • [tex]a^2=(1/2x)^2[/tex]
  • [tex]a = (1/2)x[/tex]
  • [tex]2ab = (2/3)xy[/tex]

We can work out the value of b:

  • [tex]2ab = (2/3)xy = 2(1/2x)(2/3y)[/tex]
  • [tex]b=(2/3)y[/tex]

Then we are missing b², what needs to be added:

  • [tex]b^2=(2/3y)^2=4/9y^2[/tex]

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