Question

Решите уравнение дам 10 баллов​

Answer 1

$1)\; \; (2x-1)^2=2x-1\\\\(2x-1)^2-(2x-1)=0\\\\(2x-1)\cdot (2x-1-1)=0\\\\(2x-1)\cdot (2x-2)=0\; \; \to \; \; \; x_1=\frac{1}{2}\; \; ,\; \; x_2=1\\\\\\2)\; \; (x-3)^2=4(x-3)\\\\(x-3)^2-4\, (x-3)=0\\\\(x-3)\cdot (x-3-4)=0\\\\(x-3)(x-7)=0\; \; \to \; \; \; x_1=3\; ,\; \; x_2=7\\\\\\3)\; \; 4\, (x-3)^2=(2x+6)^2\; \; ,\; \; \; 4\, (x-3)^2=2^2(x+3)^2\\\\4(x-3)^2-4(x+3)^2=0\\\\4\, \Big ((x-3)^2-(x+3)^2\Big )=0\\\\4\, \Big ((x-3)-(x+3)\Big )\Big ((x-3)+(x+3)\Big )=0\; |:4\\\\-6\cdot 2x=0\; \; \to \; \; \; x=0$

$4)\; \; (3x+4)^2=3(x+4)\\\\9x^2+24x+16=3x+12\\\\9x^2+21x+4=0\; \; ,\; \; D=297\\\\x-1=\frac{-21-\sqrt{297}}{18}\; \; ,\; \; x_2=\frac{-21+\sqrt{297}}{18}$