1) \(x^4-8x^3+11x^2+8x-12=0\)

\(\Leftrightarrow x^4-x^3-7x^3+7x^2+4x^2-4x+12x-12=0\)

\(\Leftrightarrow x^3\left(x-1\right)-7x^2\left(x-1\right)+4x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+4x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2-8x^2-8x+12x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+1\right)-8x\left(x+1\right)+12\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-2x-6x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left[x\left(x-2\right)-6\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-2=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\\x=6\end{matrix}\right.\)

Vậy ...

2) \(4x^3+x^2+x-3=0\)

\(\Leftrightarrow4x^3-3x^2+4x^2-3x+4x-3=0\)

\(\Leftrightarrow x^2\left(4x-3\right)+x\left(4x-3\right)+\left(4x-3\right)=0\)

\(\Leftrightarrow\left(4x-3\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow4x-3=0\left(\text{vì }x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\right)\)

\(\Leftrightarrow4x=3\Leftrightarrow x=\dfrac{3}{4}\)

Vậy ...

\(\left(\sqrt{2x+5}-\left(x+1\right)\right)^2+\left(\sqrt{3\left(x+1\right)}-\sqrt{x+7}\right)^2=0.\\ \)
Đến đây chắc biết phải làm gì =))
 

\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\\ \Leftrightarrow-2\sqrt{x-1}=-2\Leftrightarrow\sqrt{x-1}=1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{1}{3}\sqrt{2x}-2\sqrt{2x}+3\sqrt{2x}=12\\ \Leftrightarrow\dfrac{4}{3}\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=9\\ \Leftrightarrow2x=81\Leftrightarrow x=\dfrac{81}{2}\left(tm\right)\)

Bài 1: 

b: \(\Leftrightarrow x-2=0\)

hay x=2

b)

<=>x^3+3x^2-3x+1=0

<=> (x-1)^3=0

<=> x=1.

c)<=>x^4=x^2-2x+1

<=>x^4=(x-1)^2

<=>(x^2-x+1)(x^2+x-1)=0

Do x^2-x+1>0

=> x^2+x-1=0

<=> x=(-1+căn 5)/2; (-1-căn 5)/2

Lời giải:
ĐKXĐ: $x\geq -3,5$

PT \(\Leftrightarrow (\sqrt{2x+7}-1)+(\sqrt[3]{x+4}-1)+(x^2+8x+15)=0\)

\(\Leftrightarrow \frac{2(x+3)}{\sqrt{2x+7}+1}+\frac{x+3}{\sqrt[3]{(x+4)^2}+\sqrt[3]{x+4}+1}+(x+3)(x+5)=0\)

\(\Leftrightarrow (x+3)\left[\frac{2}{\sqrt{2x+7}+1}+\frac{1}{\sqrt[3]{(x+4)^2}+\sqrt[3]{x+4}+1}+(x+5)\right]=0\)

Với $x\geq -3,5$ dễ thấy biểu thức trong ngoặc vuông $>0$

Do đó: $x+3=0$

$\Leftrightarrow x=-3$ (thỏa mãn)

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)