Question

rozwiąż równanie (3x+2)/(x^2-9 )– (2x-7)/(x-3 )= 7/(x+3 ) POMOCY

Answer 1

( 3 x + 2)/( x^2 - 9) - ( 2 x - 7)/( x - 3) = 7 / ( x + 3)

Założenia:  x  różne od - 3 i x różnr od 3

Sprowadzamy do wspólnego mianownika:

( 3 x + 2)/ ( x^2 - 9) - [ (2x - 7)*(x + 3)]/ ( x^2 - 9) = [ 7*(x - 3)]/( x^2 - 9)

( 3 x + 2)/( x^2 - 9) - ( 2 x^2 +  6 x - 7 x - 21 )/( x^2 - 9) = ( 7 x - 21)/ ( x^2 - 9)

[ 3 x + 2 - 2 x^2  + x + 21]/ ( x^2 - 9) = ( 7 x - 21) / ( x^2 - 9)

- 2 x^2 + 4 x  + 23 = 7 x - 21

 2 x^2  + 3 x - 44 = 0

----------------------------

delta = 3^2 - 4*2*( - 44) = 9 + 352 = 361

p( delty) = 19

x = [ - 3 - 19]/4 = - 22/4 = - 5,5     lub    x = [ - 3 + 19]/4 = 16/4 = 4

Odp.   x = - 5,5  lub  x = 4

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Answer 2

Założenie x-3≠0 x+3≠0 x²-9≠0

(3x+2)/(x²-9) - (2x-7)/(x-3) = 7/(x+3)

(3x+2)/(x²-9) -((2x-7)*(x+3))/(x²-9)  -(7*(x-3))/(x²-9)=0

(3x+2-2x²-6x+7x+21-7x+21)/(x²-9)=0

(-2x²-3x+44)/(x²-9)=0  //*(x²-9)

-2x²-3x+44=0

 

Δ=9+8*44=361

√Δ=19

x₁=(3-19)/(-4)=4

x₂=(3+19)/(-4)=   -22/4

 

(x+(22/4))*(x-4)=0

x=-22/4 lub x=4