Question

resolve the 5x+6/2-x-x*2​

Answer 1

Answer:

B

Step-by-step explanation:

One critical point at x=1.. so TWO ranges

1) x<1..

x2−5x+6=2−−(x−1)x2−5x+6=2−−(x−1)..

x2−5x+6=2+x−1x2−5x+6=2+x−1..

=>x2−6x+5=0=>x2−6x+5=0 ORx2−5x−x+5=0x2−5x−x+5=0 or (x−5)(x−1)=0(x−5)(x−1)=0

x=5 or x=1..

BUT since we are looking at the range x<1, BOTH 1 and 5 are not valid..

2) x>=1

x2−5x+6=2−(x−1)x2−5x+6=2−(x−1)..

x2−5x+6=2−x+1x2−5x+6=2−x+1..

=>x2−4x+3=0=>x2−4x+3=0 OR x2−3x−x+3=0x2−3x−x+3=0 or(x−3)(x−1)=0(x−3)(x−1)=0

x=3 or x=1..

As we are looking at the range x>=1, BOTH 3 and 1 are VALID solution..

so product =1*3=3

B

Answer 2

Answer:

$\frac{(5x + 6)}{(2 - x - {x}^{2} )} \\\frac{(5x + 6)}{(2 - 2x + x - {x}^{2}) } \\ = \frac{(5x + 6)}{(2 (1 - x) + x(1 - {x}) } \\ \frac{(5x + 6)}{(1 - x) (2 + {x}) }$