Question

2) ОЧЕНЬ СРОЧНО.Розв'яжіть рiвняння пожалуйста дам 40 балов : 1) y(y+2)-7(2+y)=0
2)3x²-9x-x+3=0
3)x³-5x²+x=5​

Answer 1

1.

$y(y + 2) - 7(2 + y) = 0 \\ {y}^{2} + 2y - 14 - 7y = 0 \\ {y}^{2} - 5y - 14 = 0 \\ y_{1.2} = \frac{ - ( - 5)( + - ) \sqrt{ {( - 5)}^{2} - 4 \times 1 \times ( - 14)} }{2 \times 1} \\ y_{1.2} = \frac{ - ( - 5)( + - )9}{2} \\ y = 7 \: \: \: y = - 2$

2.

$3 {x}^{2} - 9x - x + 3 = 0 \\ 3 {x}^{2} - 10x + 3 = 0 \\ x_{1.2} = \frac{ - ( - 10)( + - ) \sqrt{ {( - 10)}^{2} - 4 \times 3 \times 3 } }{2 \times 3} \\ x_{1.2} = \frac{ - ( - 10)( + - )8}{6} \\ x = 3 \: \: \: \: x = \frac{1}{3}$

3.

${x}^{3} - 5 {x}^{2} + x = 5 \\ {x}^{3} - 5 {x}^{2} + x - 5 = 0 \\ (x - 5)( {x}^{2} + 1) = 0 \\ x - 5 = 0 \\ {x}^{2} + 1 = 0 \\ {x}^{2} = - 1 \\ x = 5 \: \: x = i \: \: x = - i$