Question

find the following products and verify the result, for x = -1, y= 2
please help me I request you please on today
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Answer 1

Step-by-step explanation:

this is my answer hope it will be helpful

Answer 2

$Given \: \Big( x^{2} - \frac{1}{2x^{2}}\big) \Big( \frac{1}{3}x^{2} - \frac{1}{x^{2}}\big)$

$=x^{2} \Big( \frac{1}{3}x^{2} - \frac{1}{x^{2}}\big)-\frac{1}{2x^{2}} \Big( \frac{1}{3}x^{2} - \frac{1}{x^{2}}\big)$

$=\frac{x^{4}}{3} + 1 - \frac{1}{6} -\frac{1}{2x^{4}}$

Therefore.,

$\red{ \Big( x^{2} - \frac{1}{2x^{2}}\big) \Big( \frac{1}{3}x^{2} - \frac{1}{x^{2}}\big)}$

$\green { = \frac{x^{4}}{3} + 1 - \frac{1}{6} -\frac{1}{2x^{4}}}$

Verification:

$If \: x = -1 \: then$

$LHS = \Big[ (-1)^{2} - \frac{1}{2(-1)^{2}}\Big][ \frac{1}{3} \times (-1)^{2} + \frac{1}{(-1)^{2}}]$

$= \Big( 1 - \frac{1}{2}\Big) \Big( \frac{1}{3} + 1 \Big)$

$= \frac{1}{2} \times \frac{4}{3}$

$= \frac{2}{3} \: --(1)$

$RHS = \frac{(-1)^{4}}{3} + 1 - \frac{1}{6} - \frac{1}{2(-1)^{4}}$

$= \frac{1}{3} + 1 - \frac{1}{6} - \frac{1}{2}$

$= \frac{ 2+6-1-3}{6}$

$=\frac{4}{6}$

$= \frac{2}{3} \: --(2)$

Therefore.,

$LHS = RHS$

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