Question

rozwiąż równania

2/x-3 + 4/x+3=

x-5/x + x/x+2=

3x/x+1 - x+2/x -2=

Answer 1

pierwsze zadanie:

 

2/x-3+4/x+3= 0 dlaczego ??    2/x-3 = -1/x  a 4/x - 3 = 1      -1+1=0

 

x-5/x  +  x/x+2=x-5+x/x=2x-5/x

3x/x+1  -  x+2/x  -2=3x/x+1  -  x+3/x  -2=2x-3/x+1-2=2x-3/x+1-2x-2/x+1=

=-5/x+1

Answer 2

2/ ( x - 3) + 4/(x + 3) = 0;  x  różne od  3  i x różne od ( -3)

[ 2*( x + 3)] / ( x^2 - 3^2) + [ 4*( x - 3)]/(x^2 - 3^2) = 0

[ 2 x + 6 + 4 x - 12 ] / (x^2 - 9) = 0

6x - 6 = 0

6 x = 6

x = 1

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( x - 5)/ x + x/( x + 2) = 0 ;  x  różne od  0  i  x różne od  (-2)

[( x - 5)*(x + 2)]/ [ x*( x + 2)] + [ x*x]/[ x*(x + 2)] = 0

[ x^2 + 2 x - 5 x - 10 ]/( x^2 + 2 x] + x^2 / [ x^2 +2 x] = 0

2 x^2 - 3 x - 10]/( x^2 + 2 x ) = 0

2 x^2 - 3 x - 10 = 0

delta = (-3)^2 - 4*2*(-10) = 9 + 80 = 89

p(delty) = p(89)

x1 = [ 3 - p(89)]/4;  x2 = [ 3 + p(89)]/4

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3x/(x + 1) - ( x + 2)/(x -2) = 0;  x różne od ( -1) i   x różne  od 2

[ 3x*(x -2)]/[ (x +1)*(x - 2)] - [ (x + 1)*(x + 2)]/[ ( x + 1)*( x - 2)] = 0

[ 3 x^2 - 6 x]/[ ( x+1)*( x -2)] - [ x^2 + 2 x + x + 2]/[ (x + 10*(x -2)] = 0

[ 3 x^2 - 6 x - x^2 -3 x - 2 ] / { ( x +1)*(x -2)] = 0

2 x^2 - 9 x - 2 = 0

delta = ( -9)^2 - 4*2*(-2) = 81 + 16 = 97

p(delty) = p(97)

x1 = [ 9 - p(97)]/4 ,  x2 = [ 9 + p(97)]/4

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p(97) - pierwiastek kwadratowy z 97