Question

HCF of ((x^(2)-9)(x-3))/(4) and (x^(2)+6x+9)/(8) is
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Answer 1

Answer:

1st exp=x2-9

        =(x+3)(x-3)

2nd exp=x2-6x+9

          =x2-3x-3x+9

             =x(x-3)-3(x-3)

            =(x-3)(x-3)

hcf=(x-3)

Step-by-step explanation:

Answer 2

Answer: $\frac{x+3}{8}$

Step-by-step explanation:

                               $HCF of Fraction =\frac{HCF of Numerator}{LCM of Denominator}$

1st expression:

                            $\frac{(x^{2} -9)(x-3)}{4}$ = $\frac{(x^{2} -3^2)(x-3)}{4}$

we know, $(a^2-b^2)=(a-b)(a+b)$

          Therefore, $\frac{(x^{2} -3^2)(x-3)}{4}$ = $\frac{(x-3)(x+3)(x-3)}{4} = \frac{(x+3)(x-3)^2}{4}$

2nd expression:

                          $\frac{x^2+6x+9}{8}$

we know, $(a+b)^2=a^2+2ab+b^2$

           Therefore, $\frac{x^2+6x+9}{8}$ = $\frac{(x+3)^2}{8}$

HCF of the Numerators = (x+3)

LCM of the Denominators:

      $4 = 2^2\\8= 2^3\\LCM = 2^3 = 8$

Answer:

                   $HCF of Fraction =\frac{HCF of Numerator}{LCM of Denominator} = \frac{x+3}{8}$