x^3 - 3x^2 + 3x - 1 = 0

( x- 1)^3    = 0 

=> x -1 = 0

=> x = 1

3/4x + 5-(2/3x-4)-(1/6+1)=(1/3x+4)-(1/3x-3)

=3/4x+5-2/3x-4-1/6x+1=1/3x+4-1/3x-3

=-1/12=7

x=84

Đ/S...

\(\left(\dfrac{3x}{4}+5\right)-\left(\dfrac{2x}{3}-4\right)-\left(\dfrac{x}{6}+1\right)=\left(\dfrac{1}{3}+4\right)-\left(\dfrac{1}{3}x-3\right)\)

\(\Leftrightarrow\dfrac{3x}{4}-\dfrac{2x}{3}-\dfrac{x}{6}+5+4-1=\dfrac{13}{3}-\dfrac{1}{3}x+9\)

\(\Leftrightarrow\dfrac{9x-8x-2x}{12}+8=\dfrac{13-x}{3}+\dfrac{27}{3}\)

\(\Leftrightarrow\dfrac{-x}{12}+\dfrac{96}{12}=\dfrac{40-x}{3}\Leftrightarrow\dfrac{96-x}{12}=\dfrac{160-4x}{12}\)

\(\Rightarrow96-160=-4x+x\Leftrightarrow-64=-3x\Leftrightarrow x=\dfrac{64}{3}\)

|2-3x|=x-1

2-3x=x-1 hoặc 2-3x=x-1

TH1:2-3x=x-1

3x-x=2-(-1)

2x=2+1

2x=3

x=3/2

TH2:2-3x=-(x-1)

2-3x=-x+1

3x+x=2-1

4x=1

x=1/4

Ta có : |2 - 3x| = x - 1

\(\Leftrightarrow\orbr{\begin{cases}2-3x=x-1\\2-3x=-x+1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-3x-x=-1-2\\-3x+x=1-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-4x=-3\\-2x=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\end{cases}}\)

là đấy

kết quả là2

|x²+|3x-1||=x²+7

Vì |x²+|3x-1|| \(\ge\) 0;

    x²+7 \(\ge\) 0

nên VT=VP

<=>x²+|3x-1|=x²+7

<=>|3x-1|=x²+7-x²=7

<=>\(\int^{3x-1=7\Rightarrow3x=8\Rightarrow x=\frac{8}{3}}_{3x-1=-7\Rightarrow3x=-6\Rightarrow x=-2}\)

Vậy x \(\in\) (-2;8/3}