Question

Решите уравнения:
1) (2x-1)²=2x-1;
2) (x-3)²=4(x-3);
3) 4(x-3)²=(2x+6)²;
4) (3x+4)²=3(x+4);​

Answer 1

$1)\ \ (2x-1)^2=(2x-1)\\\\t=2x-1\ \ \to \ \ \ t^2-t=0\ \ ,\ \ t\, (t-1)=0\ \ ,\ \ t_1=0\ ,\ t_2=1\\\\a)\ \ 2x-1=0\ \ ,\ \ 2x=1\ \ ,\ \ x=0,5\\\\b)\ \ 2x-1=1\ \ ,\ \ 2x=2\ \ ,\ \ x=1\\\\Otvet:\ \ x_1=0,5\ \ ,\ \ x_2=1\ .\\\\\\2)\ \ (x-3)^2=4(x-3)\\\\t=x-3\ \ ,\ \ \ t^2-4t=0\ \ ,\ \ \ t\, (t-4)=0\ \ ,\ \ t_1=0\ ,\ t_2=4\\\\a)\ \ x-3=0\ \ ,\ \ x=3\\\\b)\ \ x-3=4\ \ ,\ \ x=7\\\\Otvet:\ \ x_1=3\ ,\ x_2=7\ .$

$3)\ \ 4(x-3)^2=(2x+6)^2\ \ \ \to \ \ \ 4(x-3)^2=\Big(2(x+3)\Big)^2\ \ ,\\\\4(x-3)^2=4(x+3)^2\ \ ,\ \ \ (x-3)^2=(x+3)^2\ \ ,\\\\(x-3)^2-(x+3)^2=0\ \ ,\ \ \ \ \star \ a^2-b^2=(a-b)(a+b)\ \star \\\\\Big((x-3)-(x+3)\Big)\Big((x-3)+(x+3)\Big)=0\\\\(x-3-x-3)(x-3+x+3)=0\\\\-6\cdot 2x=0\ \ \ \to \ \ \ x=0\\\\Otvet:\ \ x=0\ .\\\\\\4)\ \ (3x+4)^2=3(x+4)\\\\9x^2+24x+16=3x+12\\\\9x^2+21x+4=0\ \ ,\ \ \ \ D=441-144=297\ ,\\\\Otvet:\ \ x_{1}=\dfrac{-21-\sqrt{297}}{18}\ \ ,\ \ x_2=\dfrac{-21+\sqrt{297}}{18}$