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ROZWIAZ
(2x-3)kwadrat [pod pierwiastkiem drugiego stopnia]=7 a
(roznica kwadratow)
√(2x-3)² = 7 /()²
(2x -3)² = 49
4x² - 12x +9 -49 = 0
4x² -12x -40 = 0 /:4
x² -3x -10 = 0
Δ = (-3)² - 4*1*(-10) = 9 + 40 = 49
√Δ = √49 = 7
x1 = (3 -7): 2*1 = (-4) :2 = -2
x2 = (3 +7) : 2*1 = 10 : 2 = 5
Odp x = -2 lub x = 5
zad 2
rozwiaz
(2x-3)/(x+1) -1 = x/(x-2)
/ ulamek
[(2x -3) - (x +1) ] : (x +1) = x : (x -2)
(2x -3-x -1) : (x +1) = x :( x-2)
(x -4): (x +1) = x : ( x -2)
korzystam z proporcji ,że iloczyn wyrazów skrajnych = iloczynowi wyrazów środkowwych
(x-4)*(x-2) = x*( x+1)
x² -2x -4x +8 = x² +x
x² -6x -x² -x = -8
-7x = -8
x = (-8): (-7)
x = 8/7
spr.
L(8/7) =[ 2*(8/7) -3] : ( 8/7 + 1) -1 = [ 16/7 - 21/7] : [ 8/7 + 7/7] -1=
= ( -5/7) : ( 15/7) -1 = (-5/7)*(7/15) -1 = -1/3 -1 = -4/3
P(8/7) = (8/7) : [ 8/7 -2]= (8/7) : ( 8/7 -14/7) = (8/7): (-6/7)= (8/7)*(-7/6)=
= - 8/6 = - 4/3
L(8/7) = P(8/7)