The product of the reciprocals of x - 3x - 2 and x2 -4x2 - 9 is

The correct option is D x 3x 2 The reciprocal of x + 3x + 2 is x + 2x + 3 #160;And the reciprocal of x^2 4x^2 9 is x^2 9x ...

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Standard X

Mathematics

Multiplication of Rational Expressions

The product o...

Question

The product of the reciprocals of $\frac{x + 3}{x + 2}$ and $\frac{x^{2} - 4}{x^{2} - 9}$ is

\frac{1}{(x - 3) (x - 2)}

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\frac{x - 2}{x - 3}

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\frac{x - 3}{x - 2}

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(x - 3) (x - 2)

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Solution

The correct option is D $\frac{x - 3}{x - 2}$
The reciprocal of $\frac{x + 3}{x + 2}$ is $\frac{x + 2}{x + 3}$
And the reciprocal of $\frac{x^{2} - 4}{x^{2} - 9}$ is $\frac{x^{2} - 9}{x^{2} - 4}$
$\Rightarrow \frac{x^{2} - 9}{x^{2} - 4} = \frac{(x - 3) (x + 3)}{(x - 2) (x + 2)}$
$\Rightarrow \frac{(x + 2)}{(x + 3)} \times \frac{(x - 3)}{(x - 2)} \times \frac{(x + 3)}{(x + 2)} = \frac{x - 3}{x - 2}$ .

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Similar questions

Q. The product of the reciprocal of

\frac{x + 3}{x + 2}

and

\frac{x^{2} - 4}{x^{2} - 9}

Q. Which of the following are quadratic equations?

(i) x² + 6x − 4 = 0
(ii)

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

(iii)

x^{2} + \frac{1}{x^{2}} = 5

(iv)

x - \frac{3}{x} = x^{2}

(v)

2 x^{2} - \sqrt{3 x} + 9 = 0

(vi)

x^{2} - 2 x - \sqrt{x} - 5 = 0

(vii) 3x² − 5x + 9 = x² − 7x + 3
(viii)

x + \frac{1}{x} = 1

(ix) x² − 3x = 0
(x)

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

(xi) (2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)

x + \frac{1}{x} = x^{2}, x \neq 0

(xiii) 16x² − 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)³ = x³ − 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

Q. Find the largest integral

x

which satisfies the following inequality:

\frac{x + 4}{x^{2} - 9} - \frac{2}{x + 3} < \frac{4 x}{3 x - x^{2}}

Q. If

f (x) = {\begin{matrix} x^{2} - 3 x + 2, & x < 2 x - 3, & x \geq 2 \end{matrix}

g (x) = {\begin{matrix} x - 2, & x < 3 x^{2} + x + 1, & x \geq 3 \end{matrix}

Then which of the following is/are true

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The product of the reciprocals of x+3x+2 and x2−4x2−9 is

The correct option is D x−3x−2The reciprocal of x+3x+2 is x+2x+3 And the reciprocal of x2−4x2−9 is x2−9x2−4⇒x2−9x2−4=(x−3)(x+3)(x−2)(x+2)⇒(x+2)(x+3)×(x−3)(x−2)×(x+3)(x+2)=x−3x−2.

The product of the reciprocals of $\frac{x + 3}{x + 2}$ and $\frac{x^{2} - 4}{x^{2} - 9}$ is