Question

$\frac{x-3}{x-4}\ \textgreater \ \frac{x^{2}-5x+6 }{x-2}$

Answer 1

x-2≠0, x≠2
x-4≠0, x≠4
x²-5x+6=x²-4x+4-x+2=(x-2)²-(x-2)=(x-2)(x-1)
(x²-5x+6)/(x-2)=x-1

(x-3)/(x-4)>x-1
x-3>(x-1)(x-4), pt x-4>0, x>4
x-3>x²-4x-x+4
x-3>x²-5x+4
x²-5x+4<x-3
x²-6x+7<0
Δ=36-28=8
x1,2=(6+/-2√2)/2=3+/-√2 dar x>4
deci S1=(4; 3+√2)

(x-3)/(x-4)>x-1
x-3<(x-1)(x-4), pt x-4<0, x<4
x²-6x+7>0
x1,2=(6+/-2√2)/2=3+/-√2 dar x<4
deci S2=(-∞; 3-√2)

functia este negativa intre solutii, pe intervalul (3-√2; 3+√2) si pozitiva in afara lor adica (-∞; 3-√2)∪(3+√2; ∞)
Solutia inecuatiei este
S=S1∪S2=(-∞, 3-√2)∪(4; 3+√2)