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zhenek [66]
3 years ago
9

What is the difference?

Mathematics
1 answer:
AfilCa [17]3 years ago
3 0

Answer:

Option b) is correct.

The difference of given expression is

\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{1x+1}{x^2-9}\right)=\frac{(x+2)(x+5)}{(x^3-9x)}

Step-by-step explanation:

Given expression is

\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{1x+1}{x^2-9}\right)

To find their difference

\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{x+1}{x^2-9}\right)

The expression can be written as below

\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{x+1}{x^2-9}\right)=\left(\frac{2x+5}{x (x-3)}\right)-\left(\frac{3x+5}{x(x^2-9)}\right)-\left(\frac{x+1}{x^2-9}\right)

=\left(\frac{2x+5}{x(x-3)}\right)-\left(\frac{3x+5}{x(x^2-3^2)}\right)-\left(\frac{x+1}{x^2-3^2}\right)

=\left(\frac{2x+5}{x(x-3)}\right)-\left(\frac{3x+5}{x(x+3)(x-3)}\right)-\left(\frac{x+1}{(x+3)(x-3)}\right) (using a^2-b^2=(a+b)(a-b))

=\frac{(2x+5)(x+3)-(3x+5)-(x+1)x}{x(x+3)(x-3)}

=\frac{2x^2+6x+5x+15-3x-5-x^2-x}{x(x+3)(x-3)}

=\frac{x^2+7x+10}{x(x+3)(x-3)}

=\frac{(x+2)(x+5)}{x(x+3)(x-3)}

=\frac{(x+2)(x+5)}{x(x^2-3^2)}   (using a^2-b^2=(a+b)(a-b))

=\frac{(x+2)(x+5)}{x(x^2-9)}

=\frac{(x+2)(x+5)}{(x^3-9x)}

Therefore \left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{1x+1}{x^2-9}\right)=\frac{(x+2)(x+5)}{(x^3-9x)}

Option b) is correct.

The difference of given expression is

\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{1x+1}{x^2-9}\right)=\frac{(x+2)(x+5)}{(x^3-9x)}

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