We have to express the given expression 5/(x^2-5x+6) as partial fractions.

5/(x^2-5x+6)

=> 5 / ( x^2 - 3x - 2x + 6)

=> 5/ ( x(x - 3) - 2( x - 3))

=> 5/ (x - 2)(x - 3)

=> A / (x - 2) + B/ (x...

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We have to express the given expression 5/(x^2-5x+6) as partial fractions.

5/(x^2-5x+6)

=> 5 / ( x^2 - 3x - 2x + 6)

=> 5/ ( x(x - 3) - 2( x - 3))

=> 5/ (x - 2)(x - 3)

=> A / (x - 2) + B/ (x - 3)

=> [A(x - 3) + B(x - 2)]/ (x - 2)(x -3)

Ax - 3A + Bx - 2B = 5

=> Ax + Bx - 3A - 2B = 5

Equate the coefficients of x and the numeric coefficients

=> A + B = 0 and 3A + 2B = -5

=> A = -B

Substitute in 3A + 2B = -5

=> -3B + 2B = -5

=> B = 5

A = -5

**We can write 5/(x^2-5x+6) = 5/(x - 3) - 5/(x - 2).**

We need to rewrite the fraction 5/(x^2-5x+6).

First we will simplify the denominator.

==> 5/(x^2-5x+6)= 5/(x-2)(x-3)

==> 5/(x-2)(x-3)= A/(x-2) + B/(x-3)

Now we will simplify by multiplying both sides by (x-2)(x-3)

==> 5 = A(x-3) + B(x-2)

==> 5 = Ax - 3A + Bx -2B

==> 5= (A+B)x + (-3A-2B)

==> A+B = 0 ==> A = -B

==> -3A -2B = 5

==> -3(-B) - 2B = 5

==> 3B - 2B = 5

==> B= 5

==> A = -5

**==> 5/(x^2 -5x+6) = -5/(x-2) + 5/(x-3)**