Question

x²-(2x-1)(2x+1)=2-4x
срочно нада​

Answer 1

Ответ:

$\frac{1}{3},\ 1$

Объяснение:

$x^2-(2x-1)(2x+1)=2-4x\\\\x^2-4x^2+1=2-4x\\\\x^2-4x^2+1-2+4x=0\\\\-3x^2+4x-1=0\\\\D=4^2-4\cdot(-3)\cdot(-1)=16-12=4\\\\\sqrt{D}=\sqrt{4}=2\\\\x_1=\frac{-4-2}{2\cdot(-3)}=\frac{-6}{-6}=1\\\\x_2=\frac{-4+2}{2\cdot(-3)}=\frac{-2}{-6}=\frac{1}{3}$

Answer 2

$\displaystyle\bf\\x^{2} -(2x-1)(2x+1)=2-4x\\\\x^{2} -(4x^{2} -1)=2-4x\\\\x^{2} -4x^{2} +1-2+4x=0\\\\-3x^{2} +4x-1=0\\\\3x^{2} -4x+1=0\\\\D=(-4)^{2} -4\cdot3\cdot 1=16-12=4=2^{2} \\\\\\x_{1} =\frac{4-2}{6} =\frac{2}{6} =\frac{1}{3} \\\\\\x_{2} =\frac{4+2}{6} =\frac{6}{6}=1\\\\\\Otvet \ : \ \frac{1}{3} \ \ ; \ \ 1$