Ta thấy \(\left|x+\frac{1}{8}\right|\ge0\forall x;\left|x+\frac{2}{8}\right|\ge0\forall x;\left|x+\frac{5}{8}\right|\ge0\forall x\)

\(\Rightarrow\left|x+\frac{1}{8}\right|+\left|x+\frac{2}{8}\right|+\left|x+\frac{5}{8}\right|\ge0\)

\(\Rightarrow4x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{8}+x+\frac{2}{8}+x+\frac{5}{8}=4x\)

\(\Rightarrow3x+1=4x\)

=> x = 1 (t/m)

Vậy x=1

a: \(x^3-4x^2-x+4=0\)

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

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

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

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

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

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

=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

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

3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

4x.(x+1)-8(x+1)=0

(4x-8)(x+1)=0

suy ra x=2 hoặc x=-1

1) \(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x^2+4x=8x+8\Leftrightarrow4x^2-4x-8=0\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

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

a) \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

Vậy \(x=1;-3\)

b) \(x^2-4x+8=2x-1\)

\(\Leftrightarrow x^2-4x+8-2x+1=0\)

\(\Leftrightarrow x^2-6x+9=0\)

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy x=3

a)    \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

Vậy \(x=1;-3\)

b)    \(x^2-4x+8=2x-1\)

\(\Leftrightarrow x^2-4x+8-2x+1=0\)

\(\Leftrightarrow x^2-6x+9=0\)

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Vậy \(x=3\)

\(\Rightarrow\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)

\(\Rightarrow\frac{3}{4}x-2x-\frac{1}{4}x=-6+\frac{1}{4}\)

\(\Rightarrow-\frac{3}{2}x=-\frac{23}{4}\)

\(\Rightarrow x=\frac{23}{4}:\frac{3}{2}=\frac{23}{6}\)

\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+\frac{11}{3}\)

\(\Rightarrow\frac{5}{9}x=\frac{13}{3}\)

\(\Rightarrow x=\frac{13}{3}:\frac{5}{9}=\frac{39}{5}\)

\(\Rightarrow\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)

\(\Rightarrow\frac{3}{4}x-2x-\frac{1}{4}x=-6+\frac{1}{4}\)

\(\Rightarrow-\frac{3}{2}x=-\frac{23}{4}\)

\(\Rightarrow x=\frac{23}{4}:\frac{3}{2}=\frac{23}{6}\)

\(\Rightarrow\frac{8}{9}x-\frac{1}{3}x=\frac{2}{3}+\frac{11}{3}\)

\(\Rightarrow\frac{5}{9}x=\frac{13}{3}\)

\(\Rightarrow x=\frac{13}{3}:\frac{5}{9}=\frac{39}{5}\)