f(x)=ax+b
A(1,4) ∈ graficului⇒f(1)=4, a+b=4
B(-2,8)∈ graficului⇒f(-2)=8, -2a+b=8
scadem cele doua relatii si obtinem
3a=-4
a=-4/3
b=4-a=4+4/3=12/3+4/3=16/3
b. a=2, b=3
f(x)=2x+3
f(x)/(x-1)+x=5, x-1≠0, x≠1 (este conditia pt existenta numitorului)
(2x+3)/(x-1)+x=5
(2x+3)/(x-1)+x(x-1)/(x-1)=5
(2x+3+x²-x)/(x-1)=5
(x²+x+3)/(x-1)=5
x²+x+3=5(x-1)
x²+x+3=5x-5
x²-4x+8=0
x1,x2=(4+/-√16-32)/2 ∉N
S=Ф
