Rozwiąż równania kwadratowe najprostszą metodą: 1. 5x^-7x=0 2. (2x-3)(3x-5)=(2x-3)(5x-1) 3. (2x-3)^ .... Question from @Olexandra1

December 2018 1 24 Report

Rozwiąż równania kwadratowe najprostszą metodą:
1. 5x^-7x=0
2. (2x-3)(3x-5)=(2x-3)(5x-1)
3. (2x-3)^ - (2x+1)^=0
4. 3(x+1)(x-2)=2x^-x-7
5. (4x^-5)^ =4x^4 - 9x^ + 5

^ to potega do kwadratu
a w ostatnim przykladzie ^4 to czwarta potega

Milena233 $\textbf{1.} \\ \\ 5x^2-7x=0 \\ x(5x-7)=0 \\ 5x-7=0 \ \ \ \ \ \ \vee \ \ \ \ \ x=0 \\ 5x=7 \\ x= \frac{7}{5} \\ x \in \ \left\{ 0, \frac{7}{5} \right\}$

$\textbf{2.} \\ \\ (2x-3)(3x-5)=(2x-3)(5x-1) \\ 6x^2-10x-9x+15=10x^2-2x-15x+3 \\ -4x^2-2x+12=0 \\ -2x^2-x+6=0 \\ \Delta=1-4*(-2)*6=49 \\ x_1= \frac{1-7}{-4}= \frac{3}{2} \\ x_2= \frac{1+7}{-4}=-2 \\ x \in \ \left\{ -2, \frac{3}{2} \right\}$

$\textbf{3.} \\ \\ (2x-3)^ 2- (2x+1)^2=0 \\ 4x^2-12x+9-4x^2-4x-1=0 \\ -16x+8=0 \\ -16x=-8 \\ x= \frac{1}{2}$

$\textbf{4.} \\ \\ 3(x+1)(x-2)=2x^2-x-7 \\ 3(x^2-2x+x-2)=2x^2-x-7 \\ 3x^2-3x-6=2x^2-x-7 \\ x^2-2x+1=0 \\ \Delta=4-4*1*1=0 \\ x_0= \frac{2}{2}=1$

$\textbf{5.} \\ \\ (4x^2-5)^2 =4x^4 - 9x^2 + 5 \\ 16x^4- 40x^2+25=4x^4-9x^2+5 \\ 12x^4 -31x^2+20=0 \\ t=x^2 \geq 0 \\ 12t^2-31t+20=0 \\ \Delta=961-4*12*20=1 \\ t_1= \frac{31-1}{24}= \frac{5}{4} \\ t_2= \frac{31+1}{24}= \frac{4}{3}$
$x^2= \frac{5}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vee \ \ \ \ \ \ x^2= \frac{4}{3} \\ x= \frac{ \sqrt{5} }{2} \ \ \ \vee \ \ \ x=- \frac{ \sqrt{5} }{2} \ \ \ \ \ \ \ \vee \ \ \ \ \ \ x= \frac{2}{ \sqrt{3} } = \frac{2 \sqrt{3} }{ 3 } \ \ \ \ \vee \ \ \ \ x=- \frac{2 \sqrt{3} }{ 3 } \\ \\ x \in \ \left\{-\frac{2 \sqrt{3} }{ 3 }, -\frac{ \sqrt{5} }{2}, \frac{ \sqrt{5} }{2}, \frac{2 \sqrt{3} }{ 3 } \right\}$

2 votes Thanks 1