\(x^3-5x^2+6x=0\)

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

\(\Leftrightarrow\)x=3;x=2

Vậy S={3;2}

x³-5x²+6x=0

=>x(x²-5x+6)=0

=>x=0 hoặc x²-5x+6=0

          =>x(x-5+6)=0

=>x-5+6=0

=>x-5=-6

=>x=-1

Vậy x =0 hoặc x =-1

(* là mũ 3, ^ là mũ 2 nhé)

x*- 5x^ + 6x=0

<=>x* - 3x^ - 2x^ + 6x=0

<=>(x*- 2x^)-(3x^+ 6x)=0

<=>x^(x-2)-3x(x-2)=0

<=>(x-2)(x^-3x) =0

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

<=> _x-2=0            <=> _x=2

       | x=0                      | x=0

        _x-3=0                  _x=3

vậy S= {2;0;3}

trình bày hơi phức tạp thông cảm =)))))

\(a,=\left(3x+1\right)^2-y^2=\left(3x-y+1\right)\left(3x+y+1\right)\\ b,=x\left(x^2-5x+6\right)=x\left(x^2-2x-3x+6\right)=x\left(x-2\right)\left(x-3\right)\)

1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

a: P(x)=6x^3-4x^2+4x-2

Q(x)=-5x^3-10x^2+6x+11

M(x)=x^3-14x^2+10x+9

b: \(C\left(x\right)=7x^4-4x^3-6x+9+3x^4-7x^3-5x^2-9x+12\)

=10x^4-11x^3-5x^2-15x+21

Ta có: 

Ta có:

x3+3x2-10x=0

=>x(3+3.2-10)=0

=>x=0

x3-5x2-14x=0

=>x(3-5.2-14)=0

=>x=0

x3+5x2-24x=0

=>x(3+5.2-24)=0

=>x=0

Câu a)

\(x^3+3x^2-10=0\Rightarrow x\left(x^2+3x-10\right)=0\Rightarrow x\left(x^2-2x+5x-10\right)=0\Rightarrow x\left(x\left(x-2\right)+5\left(x-2\right)\right)=0\Rightarrow x\left(x+5\right)\left(x-2\right)=0\)

\(\Rightarrow x=0;x=5;x=2\)

Câu b:

\(x^3-5x^2-14x=0\Rightarrow x\left(x^2-5x-14\right)=0\Rightarrow x\left(x^2+2x-7x-14\right)=0\Rightarrow x\left(x\left(x+2\right)-7\left(x+2\right)\right)=0\Rightarrow x\left(x-7\right)\left(x+2\right)=0\)

\(\Rightarrow x=0;x=7;x=-2\)

b: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)

c: \(\Leftrightarrow\left(x-1\right)\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\\x=-5\end{matrix}\right.\)