Giải thích các bước giải:

a.Ta có:

$4\left(x-1\right)^2\:-\:25\left(2-3x\right)^2\:=\:0$

$\to 4(x-1)^2=25(2-3x)^2$

$\to 2(x-1)=5(2-3x)$ hoặc $2(x-1)=-5(2-3x)$

$\to x=\dfrac{12}{17}$ hoặc $x=\dfrac8{13}$

b.Ta có:

$(2x+5)^2 + 4(2x+5) = -4$

$\to (2x+5)^2 + 4(2x+5) +4=0$

$\to (2x+5+2)^2=0$

$\to (2x+7)^2=0$

$\to 2x+7=0$

$\to x=-\dfrac72$

c.Ta có:

$ (x+5)^2 + 4(x+5)(x-5) + 4(x^2 - 10x + 25) = 0$

$\to  (x+5)^2 + 4(x+5)(x-5) + 4(x-5)^2= 0$

$\to (x+5+2(x-5))^2=0$

$\to (3x-5)^2=0$

$\to 3x-5=0$

$\to x=\dfrac53$

d.Ta có:

$3(4-x)^2-6(4-x)+3=0$

$\to (4-x)^2-2(4-x)+1=0$

$\to (4-x-1)^2=0$

$\to (3-x)^2=0$

$\to 3-x=0$

$\to x=3$

e.Ta có:

$(2x-3)^2+2(4x^2-9)+(2x+3)^2=0$

$\to (2x-3)^2+2(2x+3)(2x-3)+(2x+3)^2=0$

$\to (2x-3+2x+3)^2=0$

$\to (4x)^2=0$

$\to x=0$

f.Ta có:

$(x+1)^3-3(1-x^2)(x+1)+3(1-x^2)(1-x)-(1-x)^3=0$

$\to (x+1)^3-3(1-x)(1+x)(x+1)+3(1-x)(1+x)(1-x)-(1-x)^3=0$

$\to (x+1)^3-3(x+1)^2(1-x)+3(x+1)(1-x)^2-(1-x)^3=0$

$\to ((x+1)-(1-x))^3=0$

$\to (2x)^3=0$

$\to x=0$