1. 阅读与理解:
(1) 将$2x^2 - 3x - 2$进行因式分解,我们可以按下面的方法解答.
解:①竖分二次项与常数项:$2x^2 = 2x · x, -2 = (-2) × 1$.
②交叉相乘,验中项(交叉相乘后的结果相加,其结果需等于多项式中的一次项).
③横向写出两因式:$2x^2 - 3x - 2 = (2x + 1)(x - 2)$.
我们把这种用十字相乘分解因式的方法叫作十字相乘法.

(2)例:解方程:$x^2 - 3x + 2 = 0$.
解:$(x - 2)(x - 1) = 0, \therefore x - 2 = 0$或$x - 1 = 0$,
$\therefore x_1 = 2, x_2 = 1$.
请用上述方法解答下列问题.
(3)①因式分解:$x^2 - 4x + 3 = \_\_\_\_\_\_, 3x^2 - 4x - 4 = \_\_\_\_\_\_.$
②解方程:$x^2 - 5x + 4 = 0.$
③直接写出方程$2024x^2 + 2019x - 5 = 0$的解.

①$x^2 = x · x, 3 = (-1)×(-3)$,交叉相乘验证中项,$(-1)x + (-3)x = -4x$,
$\therefore x^2 - 4x + 3 = (x - 3)(x - 1)$;$3x^2 = 3x · x, -4 = 2×(-2)$,交叉相乘验证中项,$(-2) · 3x + 2x = -4x$,$\therefore 3x^2 - 4x - 4 = (3x + 2)(x - 2)$.
②$x^2 - 5x + 4 = 0$,$\therefore (x - 4)(x - 1) = 0$,$\therefore x - 4 = 0$或$x - 1 = 0$,
$\therefore x_1 = 4$,$x_2 = 1$.
③$x_1 = \frac{5}{2024}$,$x_2 = -1$.
解析：$2024x^2 + 2019x - 5 = 0$,
$\therefore (2024x - 5)(x + 1) = 0$,$\therefore 2024x - 5 = 0$或$x + 1 = 0$,$\therefore x_1 = \frac{5}{2024}$,$x_2 = -1$.

2. 阅读下列“问题”与“提示”后，将解方程的过程补充完整，求出$x$的值.
【问题】解方程：$x^2+2x+4\sqrt{x^2+2x}-5=0$.
【提示】可以用“换元法”解方程.
解：设$\sqrt{x^2+2x}=t(t≥0)$，则$x^2+2x=t^2$，原方程可化为$t^2+4t-5=0$.
(1)续解：
(2)用上面的思想方法解方程：$\frac{x^2}{x+2}+\frac{2x+4}{x^2}=3$.

(1)解得$t=-5$(舍去)或$t=1$,$\therefore \sqrt{x^2+2x}=1$,$\therefore x^2+2x-1=0$,解得$x=\sqrt{2}-1$或$x=-\sqrt{2}-1$.
(2)设$\frac{x^2}{x+2}=t$,则原方程可化为$t+\frac{2}{t}=3$,$\therefore t^2-3t+2=0$,解得$t=2$或$t=1$.当$t=1$时,$\frac{x^2}{x+2}=1$,解得$x=2$或$x=-1$,经检验,$x=-1$或$x=2$是原方程的解;当$t=2$时,$\frac{x^2}{x+2}=2$,解得$x=1+\sqrt{5}$或$x=1-\sqrt{5}$,经检验,$x=1+\sqrt{5}$或$x=1-\sqrt{5}$是原方程的解.
$\therefore$ 原方程的解为$x_1=-1$,$x_2=2$,$x_3=1+\sqrt{5}$,$x_4=1-\sqrt{5}$.