If x - 3- which of the following is equivalent to dfrac - 1 - dfrac - 1 - x-2 - -dfrac - 1 - x-3 - - -dfrac - 2x-5 - - x - 2 -5x-6 - dfrac - - x - 2 -5x-6 - 2x-5 - 2x-5- x - 2 -5x-6

Question

Toppr Admin · Accepted Answer

Given- 11x-2-1x-3-xA0- -xA0- -xA0- -xA0- -1x-3-x-2-x-3-x-2-xA0- -xA0- -xA0- -xA0- -12x-5x2-5x-6-xA0- -xA0- -xA0- -xA0- -x2-5x-62x-5Option B is correct-

If $x > 3$ , which of the following is equivalent to $\frac{1}{\frac{1}{x + 2} + \frac{1}{x + 3}}$ ?
$\frac{2 x + 5}{x^{2} + 5 x + 6}$
$\frac{x^{2} + 5 x + 6}{2 x + 5}$
$2 x + 5$
$x^{2} + 5 x + 6$

Given: $\frac{1}{\frac{1}{x + 2} + \frac{1}{x + 3}}$
$= \frac{1}{\frac{x + 3 + x + 2}{(x + 3) (x + 2)}}$
$= \frac{1}{\frac{2 x + 5}{x^{2} + 5 x + 6}}$
$= \frac{x^{2} + 5 x + 6}{2 x + 5}$
Option B is correct.

If x>3, which of the following is equivalent to 11x+2+1x+3?2x+5x2+5x+6x2+5x+62x+52x+5x2+5x+6

Given: 11x+2+1x+3 =1x+3+x+2(x+3)(x+2) =12x+5x2+5x+6 =x2+5x+62x+5Option B is correct.

If $x > 3$ , which of the following is equivalent to $\frac{1}{\frac{1}{x + 2} + \frac{1}{x + 3}}$ ?
$\frac{2 x + 5}{x^{2} + 5 x + 6}$
$\frac{x^{2} + 5 x + 6}{2 x + 5}$
$2 x + 5$
$x^{2} + 5 x + 6$

Given: $\frac{1}{\frac{1}{x + 2} + \frac{1}{x + 3}}$
$= \frac{1}{\frac{x + 3 + x + 2}{(x + 3) (x + 2)}}$
$= \frac{1}{\frac{2 x + 5}{x^{2} + 5 x + 6}}$
$= \frac{x^{2} + 5 x + 6}{2 x + 5}$
Option B is correct.