a) \(x^3=x^5\)

=> \(x^3-x^5=0\)

=> \(x^3\left(1-x^2\right)=0\)

=> \(\orbr{\begin{cases}x^3=0\\1-x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

b) \(4x\left(x+1\right)=x+1\)

=> \(4x^2+4x-x-1=0\)

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

=> \(\left(x+1\right)\left(4x-1\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{4}\end{cases}}\)

c) \(x\left(x-1\right)-2\left(1-x\right)=0\)

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

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

=> \(\left(x-1\right)\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

d) Kết quả ?

e) \(\left(x-3\right)^2+3-x=0\)

=> \(x^2-6x+9+3-x=0\)

=> \(x^2-7x+12=0\)

=> \(x^2-3x-4x+12=0\)

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

=> (x - 4)(x - 3) = 0

=> \(\orbr{\begin{cases}x=4\\x=3\end{cases}}\)

f) Tương tự

a,x=0hoac x-2=0 hoac x+2 =0

x=0 hoac x=2 hoac x=-2

\(a,\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x^3-27-x^3+4x=1\\ \Leftrightarrow4x=28\Leftrightarrow x=7\\ c,\Leftrightarrow4x^2-4x-8=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow2x^2+6x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

c: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+9x^2=0\)

hay x=1

b) 4x(2-x)+(2x+1)^2=2
    8x-4x^2+4x^2+4x+1-2=0
    (8x+4x)+(-4x^2+4x^2)+(1-2)=0
         12x + 0 -1 =0
                 12x=1
                     x=1/12
Vậy x= 1/2
c) (x-3)^3-x^2(x-9)=0
     x^3-9x^2+27x-x^3+9x^2=0
     (x^3-x^3)+(-9x^2+9x^2)+27x=0
           0 + 0 + 27x=0
                     x= 0
Vậy x=0
    

a)

TH1: \(x< \dfrac{-2}{3}\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=-x-\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(2-0,5x+x+\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(c\right)\)

TH2: \(\dfrac{-2}{3}\le x< 4\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=2-0,5x\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(2-0,5x-x-\dfrac{2}{3}=0< =>x=\dfrac{8}{9}\left(c\right)\)

TH3: \(x\ge4\)

<=> \(\left\{{}\begin{matrix}\left|0,5x-2\right|=0,5x-2\\\left|x+\dfrac{2}{3}\right|=x+\dfrac{2}{3}\end{matrix}\right.\)

PT <=> \(0,5x-2-x-\dfrac{2}{3}=0< =>x=\dfrac{-16}{3}\left(l\right)\)

KL: x \(\left\{\dfrac{-16}{3};\dfrac{8}{9}\right\}\)

b) TH1: \(x\ge-1< =>\left|x+1\right|=x+1\)

PT <=> 2x - x -1 = \(\dfrac{-1}{2}\)

<=> x = \(\dfrac{1}{2}\) (c)

TH2: x < -1 <=> \(\left|x+1\right|=-x-1\)

PT <=> 2x + x + 1 = \(\dfrac{-1}{2}\)

<=> x = \(\dfrac{-1}{2}\) (l)

KL: x \(\in\left\{\dfrac{1}{2}\right\}\)