x^3 - 3x^2 + 3x - 1 = 0

( x- 1)^3    = 0 

=> x -1 = 0

=> x = 1

Tim x biết:(-3⁵-6⁷).(3x +27)=0

                 -280179.(3x+27)=0

                               3x+27=0

                                3x=27

                                  x=9

                       Vậy x=9

a)Từ \(x\cdot2y=\dfrac{2x}{y}\Rightarrow2x=x\cdot2y^2\)

Do \(x,y\ne 0\) nên \(2=2y^2\Rightarrow y^2=1\Rightarrow y=\pm1\)

*)Xét \(y=1\Rightarrow3x-2=2x\Rightarrow x=2\)

*)Xét \(y=-1\Rightarrow3x+2=-2x\Rightarrow x=-\dfrac{2}{5}\)

b)\(\left|4x-3\right|+\left|3xy-5\right|=0\)

Dễ thấy: \(\left\{{}\begin{matrix}\left|4x-3\right|\ge0\\\left|3xy-5\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|4x-3\right|+\left|3xy-5\right|\ge0\)

Xảy ra khi \(\left\{{}\begin{matrix}\left|4x-3\right|=0\\\left|3xy-5\right|=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}4x-3=0\\3xy-5=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\3xy-5=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{20}{9}\end{matrix}\right.\)

     \(x^4-2x^3-2x^2+3x+2=0\)

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

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

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

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

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

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

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

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

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

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

Đến đây ez r