Question

Pls ans the q in image

Answer 1

Answer:

$(3x - \frac{1}{3x} ) ^{2} - (3x + \frac{1}{3x} ) \times (3x - \frac{1}{3x} ) \\ \\ \\ = > 3x - \frac{1}{3x} (3x - \frac{1}{3x} - 3x - \frac{1}{3x} ) \\ \\ \\ = > 3x - \frac{1}{3x} ( \frac{ - 2}{3x} ) \\ \\ \\ = > - 2 + \frac{2}{9 {x}^{2} } = 0 \\ \\ \\ = > - 18 {x}^{2} + 2 = 0 \\ \\ \\ = > 18 {x}^{2} = 2 \\ \\ \\ = > {x}^{2} = \frac{1}{9} \\ \\ \\ = > x = + - ( \frac{1}{3} )$

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Answer 2

The values of x are  $\frac{2}{9x^{2} } -2$..

Step-by-step explanation:

We are given,

$(3x-\frac{1}{3x}) ^{2} -(3x+\frac{1}{3x}) *(3x-\frac{1}{3x})$

and we have to simplify this and find the value of x.

• Formula used,

1. $(a+b)(a-b)=a^{2} -b^{2}$
2. $(a-b)^{2} =a^{2} +b^{2} -2ab$

• Solving given equation

For solving the equation let us give 'A' to the term which are left side to minus sign and 'B' to term which is right to minus sign as,

$(3x-\frac{1}{3x}) ^{2} -[(3x+\frac{1}{3x}) *(3x-\frac{1}{3x})]$

$A=(3x-\frac{1}{3x}) ^{2}$    and  $B=[(3x+\frac{1}{3x}) *(3x-\frac{1}{3x})]$

and now we have to find $A-B$.

Solving term 'B'

$B=[(3x+\frac{1}{3x}) *(3x-\frac{1}{3x})]$

here we can apply formula (1)

$(a+b)(a-b)=a^{2} -b^{2}$

the values of a and b for 'B' are,

$a=3x$   and   $b=\frac{1}{3x}$

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

   $=9x^{2} -\frac{1}{9x^{2} }$

we get $B=9x^{2} -\frac{1}{9x^{2} }$.

Solving term 'A'

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

we can solve this 'A' by using the formula (2),

$(a-b)^{2} =a^{2} +b^{2} -2ab$

for term A,

$a=3x$   and   $b=\frac{1}{3x}$

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

$A=(3x)^{2}+(\frac{1}{3x} )^{2} -2*3x*\frac{1}{3x}$

   $=9x^{2} +\frac{1}{9x^{2} } -2$

we get $A=9x^{2} +\frac{1}{9x^{2} } -2$

• Now finding A-B

$A-B$

$(3x-\frac{1}{3x}) ^{2} -[(3x+\frac{1}{3x}) *(3x-\frac{1}{3x})]$

$=9x^{2} +\frac{1}{9x^{2} } -2-[9x^{2} -\frac{1}{9x^{2} } ]$

$=9x^{2} +\frac{1}{9x^{2} } -2-9x^{2} +\frac{1}{9x^{2} }$

$=\frac{2}{9x^{2} } -2$

we get the answer of this question as $\frac{2}{9x^{2} } -2$.