Đáp án+Giải thích các bước giải:

Bài `4`:
a)$(1-2x^2)-4x^2=14$

$1-4x +4x^2 -4x^2 -14=0$

$1-4x-14=0$
$-4x-13=0$
$-4x=13$
$x=\frac{13}{-4}$
$x=-\frac{13}{4}$
Vậy $x=-\frac{13}{4}$
b)$(x+5)^2 +3x-15=x^2$
$(x+5)^2 +3x-15=x^2$
$x^2 +10x +25+3x-15-x^2=0$
$13x+10=0$
$13x=-10$
$x=\frac{-10}{13}$
Vậy $x=-\frac{10}{13}$

c)$25x^2 -(5x+4)^2=24$
$25x^2 -25x^2 -40x-16-24=0$

$-40x=40$
$x=\frac{40}{-40}$
$x=-1$
Vậy $x=-1$
d)$(2x-3)^2=(2x+1)^2$

$4x^2 -12x+9-4x^2 -4x-1=0$

$-12x+9-4x-1=0$
$-16x+8=0$
$-16x=-8$
$x=\frac{-8}{-16}$
$x=\frac{1}{2}$
Vậy $x=\frac{1}{2}$
e)$(3x-2)^2=(x+10)^2$
$9x^2 -12x+4-x^2 -2x-1=0$
$8x^2 -14x+3=0$

`\Rightarrow` $\triangle =(-14)^2 -4.8.3=100$

`\Leftrightarrow` $\sqrt{\triangle}=\sqrt{100}=10$
`\Rightarrow` $\left[\begin{matrix}x_1=\frac{14+10}{2.8}\\x_2=\frac{14-10}{2.8}\end{matrix}\right.$

`\Rightarrow` $\left[\begin{matrix}x=\frac{3}{2}\\x=\frac{1}{4}\end{matrix}\right.$

Vậy $x=\frac{3}{2}$ hoặc $x=\frac{1}{4}$

f)$(3x-2)^2 =9x^2 +16$
$9x^2 -12x +4-9x^2 -16=0$|
$-12x+4-16=0$
$-12x-12=0$
$-12x=12$
$x=-1$

Vậy $x=-1$