Treść zadania

Najlepsze rozwiązanie

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  ewka13 4.2.2011 (13:19)

  a)
  \begin {cases} x - y = 3 \\ xy = 4 \end {cases}
  
  \begin {cases} x = 3+y \\ (3+y)y = 4 \end {cases}
  
  \begin {cases} x = 3+y \\ y^{2} +3y -4 =0 \end {cases}
  
  Rozwiązanie II równania kwadratowego
  
  \Delta = 9+16=25
  
  \sqrt {\Delta} = 5
  
  y_{1}=\frac {-3-5} {2} = - 4
  
  y_{2}=\frac {-3+5} {2} = 1
  
  x_{1} = 3 -4= - 1
  
  x_{2} = 3+ 1 = 4
  
  rozwiązanie : x_{1}= - 1 \ i \ y_{1} = -4 \ oraz \ x_{2} =4 \ i \ y_{2} =1
  
  c)
  \begin {cases} x -y = 2 \\ 2(x+y)^{2} +x +y -1 =0 \end {cases}
  
  \begin {cases} x = 2+y \\ 2(x^{2} +2xy + y^{2}) +x + y -1=0 \end {cases}
  
  \begin {cases} x= 2 +y \\ 2x^{2} +4xy +2y^{2} +x +y -1 =0 \end {cases}
  
  \begin {cases} x = 2+y \\ 2(2+y)^{2} +4(2+y)y +2y^{2} +2+y +y -1=0 \end {cases}
  
  \begin {cases} x = 2+y \\ 2(4 +4y+y^{2}) +8y +4y^{2}+2y^{2} +2y+1 =0 \end {cases}
  
  \begin {cases} x = 2+y \\ 8+8y+2y^{2} +8y+6y^{2}+2y+1=0 \end {cases}
  
  \begin {cases} x = 2+y \\ 8y^{2} +18y +9 =0 \end {cases}
  
  ---------------------------------
  równanie kwadratowe :
  \Delta = 324 -288=36
  
  \sqrt {\Delta} =6
  
  y_{1}=\frac {-18-6} {16} = \frac {-24} {16} = - 1\frac {1} {2}
  
  y_{2} =\frac {-18+6} {16} = \frac {-12} {16} =-\frac {3} {4}
  
  x_{1} =2 - 1\frac {1} {2} = \frac {1} {2}
  
  x_{2}=2 - \frac {3} {4} = 1\frac {1} {4}
  
  odp.x_{1} =\frac {1} {2} \ i \ y_{1} = - 1\frac {1} {2}
  
  x_{2}=1\frac {1} {4} \ i \ y_{2}=-\frac {3} {4}

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