\(x^6+2x^3+1=0\)

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

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

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

\(\Leftrightarrow x=-1\)

___________

\(x\left(x-5\right)=4x-20\)

\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

_____________

\(x^4-2x^2=8-4x^2\)

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

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

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

\(\Leftrightarrow x^2=2\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

_______________

\(\left(x^3-x^2\right)-4x^2+8x-4\)

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

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

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

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

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

a: =>2x^2+9x-6x-27=0

=>x(2x+9)-3(2x+9)=0

=>(2x+9)(x-3)=0

=>x=3 hoặc x=-9/2

b: =>-10x^2+6x-5x+3=0

=>-2x(5x-3)-(5x-3)=0

=>(5x-3)(-2x-1)=0

=>x=-1/2 hoặc x=5/3

c: =>-x^3+2x^2-x^2+4=0

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

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

=>x-2=0

=>x=2

d: =>(x^3+8)-4x(x+2)=0

=>(x+2)(x^2-2x+4)-4x(x+2)=0

=>(x+2)(x^2-6x+4)=0

=>x=-2 hoặc \(x=3\pm\sqrt{5}\)

8: \(x^3-4x^2+8x-32=0\)

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

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

=>x-4=0

=>x=4

9: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

=>\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

=>\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

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

=>x(x-2)(x+3)=0

=>x∈{0;2;-3}

10: \(x^2-10x+16=0\)

=>\(x^2-2x-8x+16=0\)

=>x(x-2)-8(x-2)=0

=>(x-2)(x-8)=0

=>x=2 hoặc x=8

11: \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)

=>\(2x^2+3\left(x^2-1\right)=5x^2+5x\)

=>\(5x^2+5x=2x^2+3x^2-3\)

=>5x=-3

=>\(x=-\frac35\)

12: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=0\)

=>\(x^2-9-\left(x^2+3x-10\right)=0\)

=>\(x^2-9-x^2-3x+10=0\)

=>-3x+1=0

=>-3x=-1

=>\(x=\frac13\)

a: \(=\dfrac{x^3-3x^2-7x+x^2-3x-7}{x^2-3x-7}=x+1\)

b:\(=\dfrac{x^3+x^2+3x^2+3x+5x+5}{x+1}=x^2+3x+5\)

c:\(=\dfrac{x^3-3x^2-7x+2x^2-6x-14}{x^2-3x-7}=x+2\)

d: \(=\dfrac{x^2\left(x+5\right)+5x+25-25}{x+5}=x^2+5-\dfrac{25}{x+5}\)

1: ĐKXĐ: x<>0

\(\sqrt{x^2+x+2} + \frac{1}{x} = \frac{13-7x}{2}\)

\(\Leftrightarrow \sqrt{x^2+x+2} = \frac{13}{2} - \frac{7x}{2} - \frac{1}{x} = \frac{13x - 7x^2 - 2}{2x}\)

=>\(\left(\sqrt{x^2+x+2} - 2\right) + \left(\frac{1}{x} - 1\right) + \frac{7x - 7}{2} = 0\)

=>\(\frac{x^2+x-2}{\sqrt{x^2+x+2} + 2}+\frac{1-x}{x}+\frac{7(x-1)}{2}=0\)

=>\((x-1)\left[\frac{x+2}{\sqrt{x^2+x+2} + 2}-\frac{1}{x}+\frac{7}{2}\right]=0\)

=>x-1=0

=>x=1(nhận)

7: ĐKXĐ: \(16-x^2>0\)

=>\(x^2<16\)

=>-4<x<4

\(\frac{x^3}{\sqrt{16-x^2}} + x^2 - 16 = 0\)

\(\Leftrightarrow \frac{x^3}{\sqrt{16-x^2}} - \left(\sqrt{16-x^2}\right)^2 = 0\)

\(\Leftrightarrow x^3 - \left(\sqrt{16-x^2}\right)^3 = 0\)

\(\Leftrightarrow x^3 = \left(\sqrt{16-x^2}\right)^3\)

\(\Leftrightarrow x = \sqrt{16-x^2} \quad (x > 0)\)

\(\Leftrightarrow x^2 = 16 - x^2\) và x>0

=>\(2x^2=16\) và x>0

=>\(x^2=8\) và 0<x<4

=>\(x=2\sqrt2\)

\(a) x^3-4x^2+8x-32=(x^3-4x^2)+(8x-32)=x^2(x-4)+8(x-4)=(x^2+8)(x-4)​\)
th1 \(X^2+8\)=0

      \(X^2=-8( vô lí)\)

Th2 x-4=0

        X=4

Phương trình có tập nghiệm S=4

Ta có: \(x^3-4x^2+8x-32=0\)

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

\(\Leftrightarrow x-4=0\)

hay x=4

\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(x^3-4x^2+8x-8\)

\(=x^3-2x^2+4x-2x^2+4x-8\)

\(=\left(x^3-2x^2+4x\right)-\left(2x^2-4x+8\right)\)

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

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

a: x(2x-7)+14=4x

=>x(2x-7)-4x+14=0

=>x(2x-7)-2(2x-7)=0

=>(2x-7)(x-2)=0

=>\(\left[\begin{array}{l}2x-7=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac72\\ x=2\end{array}\right.\)

b: \(25x^3=2x\)

=>\(25x^3-2x=0\)

=>\(x\left(25x^2-2\right)=0\)

TH1: x=0

=>x=0

TH2: \(25x^2-2=0\)

=>\(x^2=\frac{2}{25}\)

=>\(\left[\begin{array}{l}x=\frac{\sqrt2}{5}\\ x=-\frac{\sqrt2}{5}\end{array}\right.\)

c: \(\left(x-5\right)^3=x^3-125\)

=>\(x^3-15x^2+75x-125=x^3-125\)

=>\(-15x^2+75x=0\)

=>-15x(x-5)=0

=>x(x-5)=0

=>\(\left[\begin{array}{l}x=0\\ x-5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=5\end{array}\right.\)

d: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

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

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

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

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

=>\(\left[\begin{array}{l}x-1=0\\ x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=2\end{array}\right.\)