Lời giải:

1. 

PT $\Leftrightarrow (x^2+5x)^2+2(x^2+5x)-24=0$

$\Leftrightarrow t^2+2t-24=0$ (đặt $x^2+5x=t$)

$\Leftrightarrow (t-4)(t+6)=0$

$\Rightarrow t-4=0$ hoặc $t+6=0$

Nếu $t-4=0\Leftrightarrow x^2+5x-4=0$

$\Leftrightarrow x=\frac{-5\pm \sqrt{41}}{2}$

Nếu $t+6=0$

$\Leftrightarrow x^2+5x+6=0$

$\Leftrightarrow (x+2)(x+3)=0\Rightarrow x=-2$ hoặc $x=-3$

2.

PT $\Leftrightarrow (x^2-4x+1)^2+2(x^2-4x+1)-3=0$

$\Leftrightarrow t^2+2t-3=0$ (đặt $x^2-4x+1=t$)

$\Leftrightarrow (t-1)(t+3)=0$

$\Rightarrow t-1=0$ hoặc $t+3=0$

Nếu $t-1=0\Leftrightarrow x^2-4x=0\Leftrightarrow x(x-4)=0$

$\Rightarrow x=0$ hoặc $x=4$

Nếu $t+3=0\Leftrightarrow x^2-4x+4=0$

$\Leftrightarrow (x-2)^2=0\Leftrightarrow x=2$

\(\Rightarrow2x\left(x+2\right)\left(x+2\right)-8x^2=2\left(x^3-8\right).\)

\(\Rightarrow\left(2x^2+4\right)\left(x+2\right)-8x^2=2x^3-8\)

\(\Rightarrow2x^3+4x^2+4x+8=2x^2-8\)

\(\Rightarrow2x^3+4x^2-2x^2+4x=-8-8\)

\(\Rightarrow2x^3+2x^2+4x=-16\)

\(\Rightarrow2x\left(x^2+x+2\right)=-16\)tự giải tiếp nhé!

\(\Rightarrow x=-2\)

2x(x+2)^2-8x^2=2(x-2)(x^2+2x+4)

2x(x^2+4x+4)-8x^2=2(x^3-8)

2x^3+8x^2+8x-8x^2=2x^3-16

2x^3-2x^3+8x^2-8x^2+8x=-16

8x=-16

x=-2 

Vay pt co tap nghiem la :S={-2}

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. 

<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)

<=>\(2x^3+8x=2x^3-16\)

<=>\(8x=-16\)

<=>\(x=-2\)

i. 

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

<=>\(2x^2-x-3=2x^2+9x-5\)

<=>10x=2

<=>\(x=\dfrac{1}{5}\)

f. 5 – (x – 6) = 4(3 – 2x)

<=>5-x+6=12-8x

<=>7x=1

<=>x=\(\dfrac{1}{7}\)

g. 7 – (2x + 4) = – (x + 4)

<=>7-2x-4=-x-4

<=>x=7

h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

<=>

<=>

<=>\(8x=-16\)

<=>x=-2

i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)

<=>\(9x+6=0\)

<=>x=\(\dfrac{-2}{3}\)

k. (x + 1)(2x – 3) = (2x – 1)(x + 5)

<=>

<=>10x=2

<=>

a: =(x-3)(2x+5)

b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)

=>(x-2)(5-x)=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

c)=\(\left(x-1\right)^3=0\)=>x=1

a) \(\left(x^2+2x+2\right)\left(x^2+2x+3\right)=0\)

<=> \(\orbr{\begin{cases}x^2+2x+2=0\\x^2+2x+3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+1\right)^2+1=0\left(vl\right)\\\left(x+1\right)^2+2=0\left(vl\right)\end{cases}}\)

=> pt vô nghiệm

b) \(\left(x+3\right)\left(x-3\right)\left(x^2-11\right)+3=2\)

<=> \(\left(x^2-9\right)\left(x^2-11\right)+1=0\)

<=> \(\left(x^2-9\right)^2-2\left(x^2-9\right)+1=0\)

<=> \(\left(x^2-9-1\right)^2=0\)

<=> \(x^2-10=0\)

<=> \(x=\pm\sqrt{10}\)

c) \(\left(x+3\right)^4+\left(x+5\right)^4=2\)

<=> \(\left(x+4-1\right)^4+\left(x+4+1\right)^4=2\)

Đặt x + 4 = a

<=> \(\left(a-1\right)^4+\left(a+1\right)^4=2\)

<=> \(a^4-4a^3+6a^2-4a+1+a^4+4a^3+6a^2+4a+1=2\)

<=> \(a^4+12a^2=0\)

<=> \(a^2\left(a^2+12\right)=0\)

<=> a = 0 (vì a² + 12 > 0)

Vậy S = {0}

Làm nốt c 

Đặt \(x+4=t\)

\(\Leftrightarrow\left(t-1\right)^4+\left(t+1\right)^4=2\)

\(\Leftrightarrow\left(t^2+2t+t\right)\left(t^2-2t+1\right)+\left(t+1\right)^4=2\)

\(\Leftrightarrow2t^4+12t^2+2=2\Leftrightarrow2t^4+12t^2=0\)

\(\Leftrightarrow2t^2\left(t^2+6\right)=0\Leftrightarrow t^2=-6\) ( vô lí )

Phương trình vô nghiệm

a); b) Do tích = 0 

=> Từng thừa số = 0 và ta nhận xét: \(x^2+2;x^2+3>0\)

=> a) \(\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

và câu b) \(\orbr{\begin{cases}x=\frac{1}{2}\\x=5\end{cases}}\)

a; *x-1=0 <=>x=1

    *2x+5=0 <=>x=-2,5

    *x²+2=0 <=> ko có x

b; tương tự a

a/ \(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)=0\)

Vì \(x^2\ge0\Rightarrow x^2+2\ge2>0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)