\(x^3-6x^2+11x-6=0\\ \Leftrightarrow\left(x^3-x^2\right)-\left(5x^2-5x\right)+\left(6x-6\right)=0\\ \Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

x³-3x²+4=0

⇔x³+x²-4x²-4x+4x+4=0

⇔(x³+x²⁾-(4x²+4x)+(4x+4)=0

⇔x²(x+1)-4x(x+1)+4(x+1)=0

⇔(x+1)(x²-4x+4)=0

⇔(x+1)(x-2)²=0

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

vậy S={-1;2}

(x³+x²)-(4x²-4x)+(4x+4)=0

=> x²(x+1)-4x(x+1)+4(x+1)=0

=> (x²-4x+4)(x+1)=0

=> (x-2)²(x+1)=0

=> \(\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\)

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

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

\(\Leftrightarrow x^2\left(x-6\right)-\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\\x=-1\end{matrix}\right.\)

Vậy ................................

Bài nào bn

x³-6x²+11x-6=0

⇔x³-x²-5x²+5x+6x-6=0

⇔(x³-x²)-(5x²-5x)+(6x-6)=0

⇔x²(x-1)-5x(x-1)+6(x-1)=0

⇔(x-1)(x²-5x+6)=0

⇔(x-1)(x²-2x-3x+6)=0

⇔(x-1)[(x²-2x)-(3x-6)]=0

⇔(x-1)[x(x-2)-3(x-2)]=0

⇔(x-1)(x-2)(x-3)=0

=>\(\left\{{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

Vậy S={1;2;3}

x³-4x²+x+6=0

⇔x³+x²-5x²-5x+6x+6=0

⇔(x³+x²)-(5x²-5x)+(6x+6)=0

⇔x²(x+1)-5x(x+1)+6(x+1)=0

⇔(x+1)(x²-5x+6)=0

⇔(x+1)(x²-2x-3x+6)=0

⇔(x+1)[(x²-2x)-(3x-6)]=0

⇔(x+1)[x(x-2)-3(x-2)]=0

⇔(x+1)(x-2)(x-3)=0

⇔\(\left[{}\begin{matrix}x+1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=3\end{matrix}\right.\)

vậy S={-1;2;3}

Dat x²+2x+2=a (a>0)

pt<=> \(\dfrac{a-1}{a}+\dfrac{a}{a+1}=\dfrac{7}{6}\)

=> \(\dfrac{\left(a-1\right)\left(a+1\right)}{a\left(a+1\right)}+\dfrac{a.a}{a\left(a+1\right)}=\dfrac{7}{6}\)

=> \(\dfrac{a^2-1}{a\left(a+1\right)}+\dfrac{a^2}{a\left(a+1\right)}=\dfrac{7}{6}\)

=> (2a²-1).6=7a(a+1)

=> 12a²-6=7a²+7a

=> 5a²-7a-6=0

\(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)

Đặt x² + 2x + 1 = t, ta có:

\(\dfrac{t}{t+1}+\dfrac{t+1}{t+2}=\dfrac{7}{6}\)

\(\Leftrightarrow\)\(\dfrac{t\left(t+2\right)}{\left(t+1\right)\left(t+2\right)}+\dfrac{\left(t+1\right)^2}{\left(t+2\right)\left(t+1\right)}=\dfrac{7}{6}\)

\(\Leftrightarrow\) \(\dfrac{t^2+2t}{t^2+3t+2}+\dfrac{t^2+2t+1}{t^2+3t+2}=\dfrac{7}{6}\)

\(\Leftrightarrow\)\(\dfrac{t^2+2t+t^2+2t+1}{t^2+3t+2}=\dfrac{7}{6}\)

\(\Leftrightarrow\)\(\dfrac{2t^2+4t+1}{t^2+3t+2}=\dfrac{7}{6}\)

\(\Leftrightarrow\)6(2t²+4t+1) = 7(t² + 3t + 2)

\(\Leftrightarrow\) 12t² + 24t + 6 = 7t² + 21t + 14

\(\Leftrightarrow\) 12t² + 24t + 6 - 7t² - 21t - 14 = 0

\(\Leftrightarrow\) 5t² + 3t - 8 = 0

\(\Leftrightarrow\) 5t² - 5t + 8t - 8 = 0

\(\Leftrightarrow\) 5t(t - 1) + 8(t - 1) = 0

\(\Leftrightarrow\) (5t + 8)(t - 1) = 0

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5t+8=0\\t-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=-\dfrac{8}{5}\\t=1\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2+2x+1=-\dfrac{8}{5}\left(vôlívì:x^2+2x+1=\left(x+1\right)^2\ge0>-\dfrac{8}{5}\right)\\x^2+2x+1=1\end{matrix}\right.\)\(\Leftrightarrow\)x² + 2x + 1 = 1

\(\Leftrightarrow\) x² + 2x = 0

\(\Leftrightarrow\)x(x + 2) = 0

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Vậy phương trình có n₀ là S={-2;0}

thưa bạn la vô số.

xin lổi là vô số